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1.4: Radical Equations - Mathematics


Bookshelves/Algebra/Book:_Advanced_Algebra_(Redden)/05:_Radical_Functions_and_Equations/5.06:_Solving_Radical_Equations

A radical equation is any equation that contains one or more radicals with a variable in the radicand. Following are some examples of radical equations, all of which will be solved in this section:

(sqrt { 2 x - 1 } = 3)(sqrt [ 3 ] { 4 x ^ { 2 } + 7 } - 2 = 0)(sqrt { x + 2 } - sqrt { x } = 1)

The squaring property of equality states that if given real numbers (a) and (b) that are equal, the equality is retained if both numbers are squared. For example.

Given (- 3 = - 3), then squaring both quantities is also a true statement: ( ( - 3 ) ^ { 2 } = ( - 3 ) ^ { 2 }) because (9 = 9::color{Cerulean}{✓} )

The converse, on the other hand, is not necessarily true,

Given (9 = 9) which could be written (( - 3 ) ^ { 2 }= ( 3 ) ^ { 2 }) does not produce an equality if the squaring operation is removed: (- 3 eq 3::color{red}{✗} )

This is important because we will use this property to solve radical equations. Because the converse of the squaring property of equality is not necessarily true, solutions to the squared equation may not be solutions to the original. Hence squaring both sides of an equation introduces the possibility of extraneous solutions, which are solutions that do not solve the original equation. For example, to find the solution to (sqrt { x } = - 5), the technique used is to square both sides of the equal sign,( {color{Cerulean}{(}} sqrt { x } {color{Cerulean}{) ^ { 2 }}} = {color{Cerulean}{ (}}5 {color{Cerulean}{ ) ^ { 2 }}} ) which produces the solution ( x = 25). However, checking this solution produces (sqrt { 25 } = 5 ) which contradicts the original problem statement, (sqrt { x } = - 5).

For this reason, answers that result from squaring both sides of an equation must ALWAYS be checked.

How to: Solve a Radical Equation.

  1. Isolate a radical. Put ONE radical on one side of the equal sign and put everything else on the other side.
  2. Eliminate the radical. Raise both sides of the equal sign to the power that matches the index on the radical. This means square both sides if it is a square root; cube both sides if it is a cube root; etc. It is this step that can introduce extraneous roots if both sides are raised to an even power!!
  3. Solve. If the equation still contains radicals, repeat steps 1 and 2. If there are no more radicals, solve the resulting equation.
  4. Check for extraneous solutions. Check each solution to confirm the value produces a true statement when substituted back into the original equation.

Example (PageIndex{1}):

Solve: (sqrt { 3 x + 1 } = 4).

Solution

(egin{aligned} sqrt { 3 x + 1 } & = 4 ( sqrt { 3 x + 1 } ) ^ { 2 } & = ( 4 ) ^ { 2 }quadcolor{Cerulean}{Square :both:sides.} 3 x + 1 & = 16 quad:::color{Cerulean}{Solve.} 3 x & = 15 x & = 5 end{aligned})

Next, we must check.

(egin{aligned} sqrt { 3 (color{OliveGreen}{ 5}color{black}{ )} + 1 } & = 4 sqrt { 15 + 1 } & = 4 sqrt { 16 } & = 4 4 & = 4::color{Cerulean}{✓} end{aligned})

The solution set is ( {5} ).

Example (PageIndex{2}):

Solve (sqrt { x - 3 } = x - 5).

Solution

(egin{aligned} sqrt { x - 3 } &= x - 5 ( sqrt { x - 3 } ) ^ { 2 } &= ( x - 5 ) ^ { 2 }quadquadquadcolor{Cerulean}{Square:both:sides.} x - 3 &= x ^ { 2 } - 10 x + 25 end{aligned})

The resulting quadratic equation can be solved by factoring.

(egin{aligned} x - 3 & = x ^ { 2 } - 10 x + 25 0 & = x ^ { 2 } - 11 x + 28 0 & = ( x - 4 ) ( x - 7 ) end{aligned})

(egin{array} { r l } { x - 4 = 0 } & { ext { or } quad x - 7 = 0 } { x = 4 } & quadquadquadquad:{ x = 7 } end{array})

Checking the solutions after squaring both sides of an equation is not optional. Use the original equation when performing the check.

( {color{Cerulean}{Check}} ext{ } x=4)( {color{Cerulean}{Check}} ext{ } x=7 )
(egin{aligned} sqrt { x - 3 } & = x - 5 sqrt { color{Cerulean}{4}color{black}{ -} 3 } & = color{Cerulean}{4}color{black}{ -} 5 sqrt { 1 } & = - 1 1 & = - 1 quad color{red}{✗} end{aligned})(egin{aligned} sqrt { x - 3 } & = x - 5 sqrt { color{Cerulean}{7}color{black}{ -} 3 } & = color{Cerulean}{7}color{black}{ -} 5 sqrt { 4 } & = 2 2 & = 2quadcolor{Cerulean}{✓} end{aligned})

After checking, you can see that (x = 4) is an extraneous solution; it does not solve the original radical equation. Disregard that answer. The solution set consequently is just ( { 7 } ).

In the previous two examples, notice that the radical is isolated on one side of the equation. Typically, this is not the case. The steps for solving radical equations involving square roots are outlined in the following example.

Example (PageIndex{3}):

Solve: (sqrt { 2 x - 1 } + 2 = x).

Solution

Step 1: Isolate the square root.

(egin{aligned} sqrt { 2 x - 1 } + 2 &= x sqrt { 2 x - 1 } &= x - 2 end{aligned})

Step 2: Square both sides.

(egin{aligned} ( sqrt { 2 x - 1 } ) ^ { 2 } &= ( x - 2 ) ^ { 2 } 2 x - 1 &= x ^ { 2 } - 4 x + 4 end{aligned})

Step 3: Solve the resulting equation.

(egin{aligned} 2 x - 1 & = x ^ { 2 } - 4 x + 4 0 & = x ^ { 2 } - 6 x + 5 0 & = ( x - 1 ) ( x - 5 ) end{aligned})

(egin{array} { r l } { x - 1 = 0 } & { ext { or } quad x - 5 = 0 } { x = 1 } & quadquadquadquad{ x = 5 } end{array})

Step 4: Check the solutions in the original equation. Squaring both sides introduces the possibility of extraneous solutions; hence the check is required.

( {color{Cerulean}{Check} } ext{ } {x=1})( {color{Cerulean}{Check} } ext{ } {x=5})
(egin{aligned} sqrt { 2 x - 1 } + 2 & = x sqrt { 2 ( color{Cerulean}{1}color{black}{ )} - 1 } + 2 & = color{Cerulean}{1} sqrt { 1 } + 2 & = 1 1 + 2 & = 1 3 & = 1 ::color{red}{✗}end{aligned})(egin{aligned} sqrt { 2 x - 1 } + 2 & = x sqrt { 2 ( color{Cerulean}{5}color{black}{ )} - 1 } + 2 & = color{Cerulean}{5} sqrt { 9 } + 2 & = 5 3 + 2 & = 5 5 & = 5::color{Cerulean}{✓} end{aligned})

After checking, we can see that (x = 1) is an extraneous solution; it does not solve the original radical equation. This leaves ( {5 } ) as the solution set.

Sometimes there is more than one solution to a radical equation.

Example (PageIndex{4}):

Solve: (2 sqrt { 2 x + 5 } - x = 4).

Solution

Begin by isolating the term with the radical.

(egin{aligned} 2 sqrt { 2 x + 5 } - x &= 4 quadquadcolor{Cerulean}{Add:x:to:both:sides.} 2 sqrt { 2 x + 5 } &= x + 4 end{aligned})

Despite the fact that the term on the left side has a coefficient, we still consider it to be isolated. Recall that terms are separated by addition or subtraction operators.

(egin{aligned} 2 sqrt { 2 x + 5 } & = x + 4 ( 2 sqrt { 2 x + 5 } ) ^ { 2 } & = ( x + 4 ) ^ { 2 } quadquadquadcolor{Cerulean}{Square:both:sides.} 4 ( 2 x + 5 ) & = x ^ { 2 } + 8 x + 16 end{aligned})

Solve the resulting quadratic equation.

(egin{aligned} 4 ( 2 x + 5 ) & = x ^ { 2 } + 8 x + 16 8 x + 20 & = x ^ { 2 } + 8 x + 16 0 & = x ^ { 2 } - 4 0 & = ( x + 2 ) ( x - 2 ) end{aligned})

(egin{array} { r l } { x + 2 = 0 } & { ext { or } x - 2 = 0 } { x = - 2 } & quadquadquad:{ x = 2 } end{array})

Since we squared both sides, we must check our solutions.

({color{Cerulean}{Check} } ext{ } {x=-2})({color{Cerulean}{Check} } ext{ }{x=2})
(egin{array} { r } { 2 sqrt { 2 x + 5 } - x = 4 } { 2 sqrt { 2 ( color{Cerulean}{- 2}color{black}{ )} + 5 } - ( color{Cerulean}{- 2}color{black}{ )} = 4 } { 2 sqrt { - 4 + 5 } + 2 = 4 } { 2 sqrt { 1 } + 2 = 4 } { 2 + 2 = 4 } { 4 = 4 }::color{Cerulean}{✓} end{array})(egin{aligned} 2 sqrt { 2 x + 5 } - x & = 4 2 sqrt { 2 (color{Cerulean}{ 2}color{black}{ )} + 5 } - ( color{Cerulean}{2}color{black}{ )} & = 4 2 sqrt { 4 + 5 } - 2 & = 4 2 sqrt { 9 } - 2 & = 4 6 - 2 & = 4 4 & = 4 ::color{Cerulean}{✓}end{aligned})

After checking, we can see that both are solutions to the original equation. The solution set is ( { pm 2 } ).

Sometimes both of the possible solutions are extraneous.

Example (PageIndex{5}):

Solve: (sqrt { 4 - 11 x } - x + 2 = 0).

Solution

(egin{aligned} sqrt { 4 - 11 x } - x + 2 & = 0quadquadquadquadquadcolor{Cerulean}{Isolate:the:radical.} sqrt { 4 - 11 x } & = x - 2 ( sqrt { 4 - 11 x } ) ^ { 2 } & = ( x - 2 ) ^ { 2 }quadquadcolor{Cerulean}{Square:both:sides.} 4 - 11 x & = x ^ { 2 } - 4 x + 4::color{Cerulean}{Solve.} 0 & = x ^ { 2 } + 7 x 0 & = x ( x + 7 ) end{aligned})

(egin{aligned} x = 0 ext { or } x + 7 & = 0 x & = - 7 end{aligned})

Since we squared both sides, we must check our solutions.

({color{Cerulean}{Check } } ext{ } {x=0})({color{Cerulean}{Check } } ext{ } {x=-7})
(egin{aligned} sqrt { 4 - 11 x } - x + 2 & = 0 sqrt { 4 - 11 ( color{Cerulean}{0}color{black}{ )} } -color{Cerulean}{ 0}color{black}{ +} 2 & = 0 sqrt { 4 } + 2 & = 0 2 + 2 & = 0 4 & = 0 ::color{red}{✗} end{aligned})( egin{aligned} sqrt { 4 - 11 x } - x + 2 &=0 sqrt { 4 - 11 ( color{Cerulean}{- 7}color{black}{ )} } - ( color{Cerulean}{- 7}color{black}{ )} + 2 &=0 sqrt { 4 + 77 } + 7 + 2 &=0 sqrt { 81 } + 9 &=0 9 + 9 &=0 18 &=0 ::color{red}{✗} end{aligned} )

Since both possible solutions are extraneous, the equation has no solution and the solution set is ( { :: } ).

The squaring property of equality extends to any positive integer power (n). Given real numbers (a) and (b), the power property of equality states: ( ext{If}:::a = b , ext { then } a ^ { n } = b ^ { n }). This, and the fact that (( sqrt [ n ] { a } ) ^ { n } = sqrt [ n ] { a ^ { n } } = a), when (a) is nonnegative, is used to solve radical equations with indices greater than (2).

Example (PageIndex{6}):

Solve (sqrt [ 3 ] { 4 x ^ { 2 } + 7 } - 2 = 0).

Solution

(egin{aligned} sqrt [ 3 ] { 4 x ^ { 2 } + 7 } - 2 & = 0quadquadcolor{Cerulean}{Isolate:the:radical.} sqrt [ 3 ] { 4 x ^ { 2 } + 7 } & = 2 left( sqrt [ 3 ] { 4 x ^ { 2 } + 7 } ight) ^ { 3 } & = ( 2 ) ^ { 3 }quadcolor{Cerulean}{Cube:both:sides.} 4 x ^ { 2 } + 7 & = 8 quadquadcolor{Cerulean}{Solve.} 4 x ^ { 2 } - 1 & = 0 ( 2 x + 1 ) ( 2 x - 1 ) & = 0 end{aligned})

(egin{array} { r l } { 2 x + 1 = 0 } & { ext { or } quad 2 x - 1 = 0 } { 2 x = - 1 } &quadquadquadquad: { 2 x = 1 } { x = - frac { 1 } { 2 } } &quadquadquadquad::; { x = frac { 1 } { 2 } } end{array})

({color{Cerulean}{Check} } ext{ } {x=-frac{1}{2}})({color{Cerulean}{Check} } ext{ } {x=frac{1}{2}})
(egin{aligned} sqrt [ 3 ] { 4 x ^ { 2 } + 7 } - 2 & = 0 sqrt [ 3 ] { 4 left( color{Cerulean}{- frac { 1 } { 2} } ight) ^ { 2 } + 7 } - 2 & = 0 sqrt [ 3 ] { 4 cdot frac { 1 } { 4 } + 7 } - 2 & = 0 sqrt [ 3 ] { 8 } - 2 & = 0 2- 2 & = 0 0 & = 0::color{Cerulean}{✓} end{aligned})(egin{aligned} sqrt [ 3 ] { 4 x ^ { 2 } + 7 } - 2 & = 0 sqrt [ 3 ] { 4 left( color{Cerulean}{frac { 1 } { 2} } ight) ^ { 2 } + 7 } - 2 & = 0 sqrt [ 3 ] { 4 cdot frac { 1 } { 4 } + 7 } - 2 & = 0 sqrt[3]{1+7}-2 &=0 sqrt [ 3 ] { 8 } - 2 & = 0 2 - 2 & = 0 0 & = 0::color{Cerulean}{✓} end{aligned})

The solution set is ( Large{ pm frac { 1 } { 2 } Large} ).

Try It (PageIndex{6})

Solve: (x - 3 sqrt { 3 x + 1 } = 3)

Answer
The solution is (33). (The other proposed solution, (x=0) was rejected. )

It may be the case that the equation has more than one term that consists of radical expressions.

Example (PageIndex{7}):

Solve: (sqrt { 5 x - 3 } = sqrt { 4 x - 1 }).

Solution

Both radicals are considered isolated on separate sides of the equation.

(egin{aligned} sqrt { 5 x - 3 } & = sqrt { 4 x - 1 } ( sqrt { 5 x - 3 } ) ^ { 2 } & = ( sqrt { 4 x - 1 } ) ^ { 2 } quadcolor{Cerulean}{Square:both:sides.} 5 x - 3 & = 4 x - 1 quadquadquadcolor{Cerulean}{Solve.} x & = 2 end{aligned})

Check (x=2).

(egin{aligned} sqrt { 5 x - 3 } & = sqrt { 4 x - 1 } sqrt { 5 ( color{OliveGreen}{2}color{black}{ )} - 3 } & = sqrt { 4 ( color{OliveGreen}{2}color{black}{ )} - 1 } sqrt { 10 - 3 } & = sqrt { 8 - 1 } sqrt { 7 } & = sqrt { 7 }::color{Cerulean}{✓} end{aligned})

The solution set is ( { 2 } ).

Example (PageIndex{8}):

Solve: (sqrt [ 3 ] { x ^ { 2 } + x - 14 } = sqrt [ 3 ] { x + 50 }).

Solution

Eliminate the radicals by cubing both sides.

(egin{aligned} sqrt [ 3 ] { x ^ { 2 } + x - 14 } & = sqrt [ 3 ] { x + 50 } left( sqrt [ 3 ] { x ^ { 2 } + x - 14 } ight) ^ { 3 } & = ( sqrt [ 3 ] { x + 50 } ) ^ { 3 }quadcolor{Cerulean}{Cube:both:sides.} x ^ { 2 } + x - 14 & = x + 50 quadquadquadcolor{Cerulean}{Solve.} x ^ { 2 } - 64 & = 0 ( x + 8 ) ( x - 8 ) & = 0 end{aligned})

(egin{array} { r l } { x + 8 = 0 } & { ext { or } quad x - 8 = 0 } { x = - 8 } & quadquadquadquad{ x = 8 } end{array})

({color{Cerulean}{Check} } ext{ } {x=-8})( {color{Cerulean}{Check}} ext{ } {x=8} )
(egin{aligned} sqrt [ 3 ] { x ^ { 2 } + x - 14 } & = sqrt [ 3 ] { x + 50 } sqrt [ 3 ] { ( color{Cerulean}{- 8}color{black}{ )} ^ { 2 } + ( color{Cerulean}{- 8}color{black}{ )} - 14 } & = sqrt [ 3 ] { ( color{Cerulean}{- 8}color{black}{ )} + 50 } sqrt [ 3 ] { 64 - 8 - 14 } & = sqrt [ 3 ] { 42 } sqrt [ 3 ] { 42 } & = sqrt [ 3 ] { 42 }::color{Cerulean}{✓} end{aligned})egin{aligned} sqrt [ 3 ] { x ^ { 2 } + x - 14 } & = sqrt [ 3 ] { x + 50 } sqrt [ 3 ] { ( color{Cerulean}{ 8}color{black}{ )} ^ { 2 } + ( color{Cerulean}{ 8}color{black}{ )} - 14 } & = sqrt [ 3 ] { ( color{Cerulean}{8}color{black}{ )} + 50 } sqrt [ 3 ] { 64 + 8 - 14 } & = sqrt [ 3 ] { 58 } sqrt [ 3 ] { 58 } & = sqrt [ 3 ] { 58 }::color{Cerulean}{✓} end{aligned}

Answer:

The solution set is ( { pm 8 } ).

It may not be possible to isolate a radical on both sides of the equation. When this is the case, isolate the radicals, one at a time, and apply the squaring property of equality multiple times until only a polynomial remains.

Example (PageIndex{9}):

Solve: (sqrt { x + 2 } - sqrt { x } = 1)

Solution

Isolate one of the radicals.

(egin{aligned} sqrt { x + 2 } - sqrt { x } & = 1 sqrt { x + 2 } & = sqrt { x } + 1 end{aligned})

Square both sides. Be careful to apply the distributive property correctly to the right side.

(egin{aligned} ( sqrt { x + 2 } ) ^ { 2 } & = ( sqrt { x } + 1 ) ^ { 2 } x + 2 & = ( sqrt { x } + 1 ) ( sqrt { x } + 1 ) x + 2 & = sqrt { x ^ { 2 } } + sqrt { x } + sqrt { x } + 1 x + 2 & = x + 2 sqrt { x } + 1 end{aligned})

Now the equation contains only one radical. Isolate it and square both sides again.

(egin{aligned} x + 2 & = x + 2 sqrt { x } + 1 1 & = 2 sqrt { x } ( 1 ) ^ { 2 } & = ( 2 sqrt { x } ) ^ { 2 } 1 & = 4 x frac { 1 } { 4 } & = x end{aligned})

Check to see if (x = frac { 1 } { 4 }) satisfies the original equation (sqrt { x + 2 } - sqrt { x } = 1)

(egin{array} { r } { sqrt { color{OliveGreen}{frac { 1 } { 4 }}color{black}{ +} 2 } - sqrt { color{OliveGreen}{frac { 1 } { 4 }} } color{black}{=} 1 } { sqrt { frac { 9 } { 4 } } - frac { 1 } { 2 } = 1 } { frac { 3 } { 2 } - frac { 1 } { 2 } = 1 } { frac { 2 } { 2 } = 1 } { 1 = 1 }color{Cerulean}{✓} end{array})

The solution set is ( large{ frac { 1 } { 4 } large} ).

(color{Cerulean}{(}color{black}{ sqrt { x + 2 }}color{Cerulean}{ ) ^ { 2 } }color{black}{-}color{Cerulean}{ (}color{black}{ sqrt { x }}color{Cerulean}{ ) ^ { 2 }}color{black}{ =}color{Cerulean}{ (}color{black}{ 1}color{Cerulean}{ ) ^ { 2 }}color{red}{Incorrect!})

This is a common mistake and leads to an incorrect result. When squaring both sides of an equation with multiple terms, we must take care to apply the distributive property correctly.

Example (PageIndex{10}):

Solve: (sqrt { 2 x + 10 } - sqrt { x + 6 } = 1)

Solution

Isolate one of the radicals.

(egin{aligned} sqrt { 2 x + 10 } - sqrt { x + 6 } &= 1 sqrt { 2 x + 10 } & = sqrt { x + 6 } + 1 end{aligned})

Square both sides. Take care to apply the distributive property CORRECTLY to the right side.

(egin{aligned} ( sqrt { 2 x + 10 } ) ^ { 2 } & = ( sqrt { x + 6 } + 1 ) ^ { 2 } 2 x + 10 & = x + 6 + 2 sqrt { x + 6 } + 1 2 x + 10 & = x + 7 + 2 sqrt { x + 6 } end{aligned})

At this point we have one term that contains a radical. Isolate it and square both sides again.

(egin{aligned} 2 x + 10 & = x + 7 + 2 sqrt { x + 6 } x + 3 & = 2 sqrt { x + 6 } ( x + 3 ) ^ { 2 } & = ( 2 sqrt { x + 6 } ) ^ { 2 } x ^ { 2 } + 6 x + 9 & = 4 ( x + 6 ) x ^ { 2 } + 6 x + 9 & = 4 x + 24 x ^ { 2 } + 2 x - 15 & = 0 (x - 3 ) ( x + 5 ) & = 0 end{aligned})

(egin{array} { r l } { x - 3 = 0 } & { ext { or } quad x + 5 = 0 } { x = 3 } & quadquadquadquad:{ x = - 5 } end{array})

({color{Cerulean}{Check } } ext{ } {x=3})({color{Cerulean}{Check } } ext{ } {x=-5})
(egin{aligned} sqrt { 2 x + 10 } - sqrt { x + 6 } & = 1 sqrt { 2 ( color{Cerulean}{3}color{black}{ )} + 10 } - sqrt { color{Cerulean}{3}color{black}{ +} 6 } & = 1 sqrt { 16 } - sqrt { 9 } & = 1 4 - 3 & = 1 1 & = 1::color{Cerulean}{✓} end{aligned})(egin{aligned} sqrt { 2 x + 10 } - sqrt { x + 6 } &= 1 sqrt { 2 (color{Cerulean}{ - 5}color{black}{ )} + 10 } - sqrt { color{Cerulean}{- 5}color{black}{ +} 6 } &=1 sqrt { 0 } - sqrt { 1 }& =1 0 - 1 &=1 - 1 &=1::color{red}{✗} end{aligned})

The solution set is ( {3}).

Try It (PageIndex{10})

Solve: (sqrt { 4 x + 21 } - sqrt { 2 x + 22 } = 1)

Answer
The solution is (7) (The other proposed solution, (x=-3) was rejected. )

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Radical Functions and Equations Maintaining Mathematical Proficiency

Evaluate the expression.
Question 1.
7(sqrt<25>) + 10
Answer:

Question 4.
-2(3(sqrt<9>) + 13)
Answer:

Graph f and g. Describe the transformations from the graph of f to the graph of g.
Question 5.
f(x) = x g(x) = 2x – 2
Answer:

Question 6.
f(x) = x g(x) = (frac<1><3>)x + 5
Answer:

Question 7.
f(x) = x g(x) = -x + 3
Answer:

Question 8.
ABSTRACT REASONING
Let a and b represent constants, where b ≥ 0. Describe the transformations from the graph of m(x) = ax + b to the graph of n(x) = -2ax – 4b.
Answer:

Radical Functions and Equations Mathematical Practices

Mathematically proficient students distinguish correct reasoning from flawed reasoning.

Monitoring Progress

Question 1.
Which of the following square roots are rational numbers? Explain your reasoning.
(sqrt<0>, sqrt<1>, sqrt<3>, sqrt<4>, sqrt<5>, sqrt<6>, sqrt<7>, sqrt<8>, sqrt<9>)
Answer:

Question 2.
The sequence of steps shown appears to prove that 1 = 0. What is wrong with this argument?

Answer:

Lesson 10.1 Graphing Square Root Functions

Essential Question What are some of the characteristics of the graph of a square root function?

EXPLORATION 1

Graphing Square Root Functions
Work with a partner.

  • Make a table of values for each function.
  • Use the table to sketch the graph of each function.
  • Describe the domain of each function.
  • Describe the range of each function.


Answer:

EXPLORATION 2

Writing Square Root Functions
Work with a partner. Write a square root function, y = f (x), that has the given values. Then use the function to complete the table.

Answer:

Communicate Your Answer

Question 3.
What are some of the characteristics of the graph of a square root function?
Answer:

Question 4.
Graph each function. Then compare the graph to the graph of f(x) = (sqrt).
a. g(x) = (sqrt)
b. g(x) = (sqrt)
c. g(x) = 2(sqrt)
d. g(x) = -2 (sqrt)
Answer:

Monitoring Progress

Describe the domain of the function.
Question 1.
f(x) = 10 (sqrt)
Answer:

Question 2.
y = (sqrt<2x>) + 7
Answer:

Graph the function. Describe the range.
Question 4.
g(x) = (sqrt) – 4
Answer:

Question 5.
y = (sqrt<2x>) + 5
Answer:

Question 6.
n(x) = 5(sqrt)
Answer:

Graph the function. Compare the graph to the graph of f(x) = (sqrt) .
Question 7.
h(x) = (sqrt <4>x>)
Answer:

Question 9.
m(x) = -3(sqrt)
Answer:

Question 10.
Let g(x) = (frac<1> <2>sqrt+1). Describe the transformations from the graph of f(x) = (sqrt) to the graph of g. Then graph g.
Answer:

Question 11.
In Example 5, compare the velocities by finding and interpreting their average rates of change over the interval d = 30 to d = 40.
Answer:

Question 12.
WHAT IF?
At what depth does the velocity of the tsunami exceed 100 meters per second?
Answer:

Graphing Square Root Functions 10.1 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A ________ is a function that contains a radical expression with the independent variable in the radicand.
Answer:

Question 2.
VOCABULARY
Is y = 2x(sqrt<5>) a square root function? Explain.
Answer:

Question 3.
WRITING
How do you describe the domain of a square root function?
Answer:

Question 4.
REASONING
Is the graph of g(x) = 1.25(sqrt) a vertical stretch or a vertical shrink of the graph of f(x) = (sqrt)? Explain.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–14, describe the domain of the function.
Question 5.
y = 8(sqrt)
Answer:

Question 6.
y = (sqrt<4x>)
Answer:

Question 7.
y = 4 + (sqrt<-x>)
Answer:

Question 9.
h(x) = (sqrt)
Answer:

Question 10.
p(x) = (sqrt)
Answer:

Question 11.
f(x) = (sqrt<-x+8>)
Answer:

Question 13.
m(x) = 2(sqrt)
Answer:

Question 14.
n(x) = (frac<1> <2>sqrt<-x>-2)
Answer:

In Exercises 15–18, match the function with its graph. Describe the range.
Question 15.
y = (sqrt)
Answer:

Question 16.
y = 3(sqrt)
Answer:

Question 17.
y = (sqrt) – 3
Answer:

Question 18.
y = (sqrt<-x+3>)
Answer:

In Exercises 19–26, graph the function. Describe the range.
Question 19.
y = (sqrt<3x>)
Answer:

Question 20.
y = 4(sqrt<-x>)
Answer:

Question 21.
y = (sqrt) + 5
Answer:

Question 22.
y = -2 + (sqrt)
Answer:

Question 24.
g(x) = (sqrt)
Answer:

Question 25.
h(x) = (sqrt) – 2
Answer:

In Exercises 27–34, graph the function. Compare the graph to the graph of f (x) = (sqrt).
Question 27.
g(x) = (frac<1> <4>sqrt)
Answer:

Question 28.
r(x) = (sqrt<2x>)
Answer:

Question 29.
h(x) = (sqrt)
Answer:

Question 30.
q(x) = (sqrt) + 8
Answer:

Question 31.
p(x) = (sqrt<-frac<1> <3>x>)x
Answer:

Question 32.
g(x) = -5(sqrt)
Answer:

Question 33.
m(x) = –(sqrt) – 6
Answer:

Question 35.
ERROR ANALYSIS
Describe and correct the error in graphing the function y = (sqrt) + 1 .

Answer:

Question 36.
ERROR ANALYSIS
Describe and correct the error in comparing the graph of g(x) = (-frac<1> <4>sqrt) to the graph of f (x) = (sqrt).

Answer:

In Exercises 37–44, describe the transformations from the graph of f (x) = (sqrt) to the graph of h. Then graph h.
Question 37.
h(x) = 4(sqrt) – 1
Answer:

Question 38.
h(x) = (frac<1> <2>sqrt)+ 3
Answer:

Question 39.
h(x) = 2(sqrt<-x>) – 6
Answer:

Question 41.
h(x) = (frac<1> <3>sqrt) + 3
Answer:

Question 42.
h(x) = 2(sqrt) + 4
Answer:

Question 43.
h(x) = -2(sqrt) + 5
Answer:

Question 45.
COMPARING FUNCTIONS
The model S(d ) = (sqrt<30df>) represents the speed S (in miles per hour) of a van before it skids to a stop, where f is the drag factor of the road surface and d is the length (in feet) of the skid marks. The drag factor of Road Surface A is 0.75. The graph shows the speed of the van on Road Surface B. Compare the speeds by finding and interpreting their average rates of change over the interval d = 0 to d = 15.

Answer:

Question 46.
COMPARING FUNCTIONS
The velocity v (in meters per second) of an object in motion is given by v(E ) = (sqrt>), where E is the kinetic energy of the object (in joules) and m is the mass of the object (in kilograms). The mass of Object A is 4 kilograms. The graph shows the velocity of Object B. Compare the velocities of the objects by finding and interpreting the average rates of change over the interval E = 0 to E = 6.

Answer:

Question 47.
OPEN-ENDED
Consider the graph of y = (sqrt).
a. Write a function that is a vertical translation of the graph of y = (sqrt).
b. Write a function that is a reflection of the graph of y = (sqrt).
Answer:

Question 48.
REASONING
Can the domain of a square root function include negative numbers? Can the range include negative numbers? Explain your reasoning.
Answer:

Question 49.
PROBLEM SOLVING
The nozzle pressure of a fire hose allows firefighters to control the amount of water they spray on a fire. The flow rate f(in gallons per minute) can be modeled by the function f = 12(sqrt

), where p is the nozzle pressure (in pounds per square inch).

a. Use a graphing calculator to graph the function. At what pressure does the flow rate exceed 300 gallons per minute?
b. What happens to the average rate of change of the flow rate as the pressure increases?
Answer:

Question 50
PROBLEM SOLVING
The speed s (in meters per second) of a long jumper before jumping can be modeled by the function s = 10.9(sqrt), where h is the maximum height (in meters from the ground) of the jumper.

a. Use a graphing calculator to graph the function. A jumper is running 9.2 meters per second. Estimate the maximum height of the jumper.
b. Suppose the runway and pit are raised on a platform slightly higher than the ground. How would the graph of the function be transformed?
Answer:

Question 51.
MATHEMATICAL CONNECTIONS
The radius r of a circle is given by r = (sqrt>), where A is the area of the circle.

a. Describe the domain of the function. Use a graphing calculator to graph the function.
b. Use the trace feature to approximate the area of a circle with a radius of 5.4 inches.
Answer:

Question 52.
REASONING
Consider the function f(x) = 8a(sqrt).
a. For what value of a will the graph of f be identical to the graph of the parent square root function?
b. For what values of a will the graph of f be a vertical stretch of the graph of the parent square root function?
c. For what values of a will the graph of f be a vertical shrink and a reflection of the graph of the parent square root function?
Answer:

Question 53.
REASONING
The graph represents the function f(x) = (sqrt).

a. What is the minimum value of the function?
b. Does the function have a maximum value? Explain.
c. Write a square root function that has a maximum value. Does the function have a minimum value? Explain.
d. Write a square root function that has a minimum value of -4.
Answer:

Question 54.
HOW DO YOU SEE IT?
Match each function with its graph. Explain your reasoning.

Answer:

Question 55.
REASONING
Without graphing, determine which function’s graph rises more steeply, f(x) = 5(sqrt) or g(x) = (sqrt<5x>). Explain your reasoning.
Answer:

Question 56.
THOUGHT PROVOKING
Use a graphical approach to find the solutions of x – 1 = (sqrt<5x-9>). Show your work. Verify your solutions algebraically.
Answer:

Question 57.
OPEN-ENDED
Write a radical function that has a domain of all real numbers greater than or equal to -5 and a range of all real numbers less than or equal to 3.
Answer:

Maintaining Mathematical Proficiency

Evaluate the expression.(Section 6.2)
Question 58.
(sqrt [ 3]< 343 >)
Answer:

Question 59.
(sqrt [ 3]< -64 >)
Answer:

Factor the polynomial.(Section 7.5)
Question 61.
x 2 + 7x + 6
Answer:

Question 62.
d 2 – 11d + 28
Answer:

Question 63.
y 2 – 3y – 40
Answer:

Lesson 10.2 Graphing Cube Roots Functions

Essential Question What are some of the characteristics of the graph of a cube root function?

EXPLORATION 1

Graphing Cube Root Functions
Work with a partner.

  • Make a table of values for each function. Use positive and negative values of x.
  • Use the table to sketch the graph of each function.• Describe the domain of each function.
  • Describe the range of each function.


Answer:

EXPLORATION 2

Writing Cube Root Functions
Work with a partner. Write a cube root function, y = f(x), that has the given values. Then use the function to complete the table.

Answer:

Communicate Your Answer

Question 3.
What are some of the characteristics of the graph of a cube root function?
Answer:

Question 4.
Graph each function. Then compare the graph to the graph of f (x) = (sqrt [ 3]< x >) .
a. g(x) = (sqrt [ 3]< x-1 >)
b. g(x) = (sqrt [ 3]< x-1 >)
c. g(x) = 2 (sqrt [ 3]< x >)
d. g(x) = -2(sqrt [ 3]< x >)
Answer:

Monitoring Progress

Graph the function. Compare the graph to the graph of f(x) = (sqrt [ 3]< x >).
Question 1.
h(x) = (sqrt [ 3]< x >) + 3
Answer:

Question 2.
m(x) = (sqrt [ 3]< x >) – 5
Answer:

Question 3.
g(x) = 4(sqrt [ 3]< x >)
Answer:

Graph the function. Compare the graph to the graph of f(x) = (sqrt [ 3]< x >).
Question 4.
g(x) =(sqrt [3]< 0.5x+5 >) + 5
Answer:

Question 5.
h(x) = 4(sqrt [3]< x >) – 1
Answer:

Question 6.
n(x) = (sqrt [ 3]< 4-x >)
Answer:

Question 7.
Let g(x) = (-frac<1> <2>sqrt[3]) – 4. Describe the transformations from the graph of f (x) = (sqrt [ 3]< x >) to the graph of g. Then graph g.
Answer:

Question 8.
In Example 4, compare the average rates of change over the interval x = 2 to x = 10.
Answer:

Question 9.
WHAT IF?
Estimate the age of an elephant whose shoulder height is 175 centimeters.
Answer:

Graphing Cube Roots Functions 10.2 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The __________ of the radical in a cube root function is 3.
Answer:

Question 2.
WRITING
Describe the domain and range of the function f(x) = (sqrt [3]< x-4 >) + 1.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, match the function with its graph.
Question 3.
y = 3(sqrt [3]< x+2 >)
Answer:

Question 4.
y = 3(sqrt [3]< x-2 >)
Answer:

Question 5.
y = 3(sqrt [3]< x+2 >)
Answer:

Question 6.
y = (sqrt [3]< x >) – 2
Answer:

In Exercises 7–12, graph the function. Compare the graph to the graph of f(x) = (sqrt [ 3]< x >).
Question 7.
h(x) = (sqrt [3]< x-4 >)
Answer:

Question 8.
g(x) = (sqrt [3]< x+1 >)
Answer:

Question 9.
m(x) = (sqrt [3]< x+5 >)
Answer:

Question 10.
q(x) = (sqrt [3]< x >) – 3
Answer:

Question 11.
p(x) = 6(sqrt [3]< x >)
Answer:

In Exercises 13–16, compare the graphs. Find the value of h, k, or a.
Question 13.

Answer:

Question 14.

Answer:

Question 15.

Answer:

Question 16.

Answer:

In Exercises 17–26, graph the function. Compare the graph to the graph of f(x) = (sqrt [3]< x >).
Question 17.
r(x) = – (sqrt [3]< x-2 >)
Answer:

Question 19.
k(x) = 5(sqrt [3]< x+1 >)
Answer:

Question 20.
j(x) = 0.5(sqrt [3]< x-4 >)
Answer:

Question 21.
g(x) = 4(sqrt [3]< x >) – 3
Answer:

Question 22.
m(x) = 3(sqrt [3]< x >) + 7
Answer:

Question 23.
n(x) = (sqrt [3]< -8x >) – 1
Answer:

Question 24.
v(x) =(sqrt [3]< 5x >) + 2
Answer:

Question 25.
q(x) = (sqrt[3]<2(x+3)>)
Answer:

In Exercises 27–32, describe the transformations from the graph of f(x) = (sqrt [3]< x >) to the graph of the given function. Then graph the given function.
Question 27.
g(x) = (sqrt [3]< x-4 >) + 2
Answer:

Question 29.
j(x) = -5(sqrt [3]< x+3 >) + 2
Answer:

Question 30.
k(x) = 6(sqrt [3]< x-9 >) – 5
Answer:

Question 31.
v(x) = (frac<1> <3>sqrt[3]) + 7
Answer:

Question 33.
ERROR ANALYSIS
Describe and correct the error in graphing the function f(x) = (sqrt [3]< x-3 >).

Answer:

Question 34.
ERROR ANALYSIS
Describe and correct the error in graphing the function h(x) = (sqrt [3]< x >) + 1.

Answer:

Question 35.
COMPARING FUNCTIONS
The graph of cube root function q is shown. Compare the average rate of change of q to the average rate of change of f(x) = 3(sqrt [3]< x >) over the interval x = 0 to x = 6.

Answer:

Question 36.
COMPARING FUNCTIONS
The graphs of two cube root functions are shown. Compare the average rates of change of the two functions over the interval x = -2 to x = 2.

Answer:

Question 37.
MODELING WITH MATHEMATICS
For a drag race car that weighs 1600 kilograms, the velocity v (in kilometers per hour) reached by the end of a drag race can be modeled by the function v = 23.8(sqrt [3]< p >), where p is the car’s power (in horsepower). Use a graphing calculator to graph the function. Estimate the power of a 1600-kilogram car that reaches a velocity of 220 kilometers per hour.
Answer:

Question 38.
MODELING WITH MATHEMATICS
The radius r of a sphere is given by the function r = (sqrt[3]<4 pi>> V), where V is the volume of the sphere. Use a graphing calculator to graph the function. Estimate the volume of a spherical beach ball with a radius of 13 inches.
Answer:

Question 39.
MAKING AN ARGUMENT
Your friend says that all cube root functions are odd functions. Is your friend correct? Explain.
Answer:

Question 40.
HOW DO YOU SEE IT?
The graph represents the cube root function f(x) = (sqrt [3]< x >).

a. On what interval is f negative? positive?
b. On what interval, if any, is f decreasing? increasing?
c. Does f have a maximum or minimum value? Explain.
d. Find the average rate of change of f over the interval x = -1 to x = 1.
Answer:

Question 41.
PROBLEM SOLVING
Write a cube root function that passes through the point (3, 4) and has an average rate of change of -1 over the interval x = -5 to x = 2.
Answer:

Question 42.
THOUGHT PROVOKING
Write the cube root function represented by the graph. Use a graphing calculator to check your answer.

Answer:

Maintaining Mathematical Proficiency

Factor the polynomial.(Section 7.6)
Question 43.
3x 2 + 12x – 36
Answer:

Question 44.
2x 2 – 11x + 9
Answer:

Question 45.
4x 2 + 7x – 15
Answer:

Solve the equation using square roots.(Section 9.3)
Question 46.
x 2 – 36 = 0
Answer:

Question 47.
5x 2 + 20 = 0
Answer:

Question 48.
(x + 4) 2 – 81
Answer:

Question 49.
25(x – 2) 2 = 9
Answer:

Radical Functions and Equations Study Skills: Making Note Cards

10.1–10.2 What Did YouLearn?

Core Vocabulary
square root function, p. 544
radical function, p. 545
cube root function, p. 552

Core Concepts
Lesson 10.1
Square Root Functions, p. 544
Transformations of Square Root Functions, p. 545
Comparing Square Root Functions Using Average Rates of Change, p. 546

Lesson 10.2
Cube Root Functions, p. 552
Comparing Cube Root Functions
Using Average Rates of Change, p. 554

Mathematical Practices

Question 1.
In Exercise 45 on page 549, what information are you given? What relationships are present? What is your goal?
Answer:

Question 2.
What units of measure did you use in your answer to Exercise 38 on page 556? Explain your reasoning.
Answer:

Study Skills: Making Note Cards

Invest in three different colors of note cards. Use one color for each of the following: vocabulary words, rules, and calculator keystrokes.

  • Using the first color of note cards, write a vocabulary word on one side of a card. On the other side, write the definition and an example. If possible, put the definition in your own words.
  • Using the second color of note cards, write a rule on one side of a card. On the other side, write an explanation and an example.
  • Using the third color of note cards, write a calculation on one side of a card. On the other side, write the keystrokes required to perform the calculation.
    Use the note cards as references while completing your homework. Quiz yourself once a day.

Radical Functions and Equations 10.1–10.2 Quiz

Describe the domain of the function.(Lesson 10.1)
Question 1.
y = (sqrt)
Answer:

Question 2.
f(x) = 15(sqrt)
Answer:

Question 3.
y = (sqrt<3-x>)
Answer:

Graph the function. Describe the range. Compare the graph to the graph of f(x) = (sqrt). (Lesson 10.1)
Question 4.
g(x) = (sqrt) + 5
Answer:

Graph the function. Compare the graph to the graph of f(x) = (sqrt [3]< x >). (Lesson 10.2)
Question 7.
b(x) = (sqrt [3]< x+2 >)
Answer:

Question 8.
h(x) = -3(sqrt [3]< x-6 >)
Answer:

Compare the graphs. Find the value of h, k, or a. (Lesson 10.1 and Lesson 10.2)
Question 10.

Answer:

Question 11.

Answer:

Question 12.

Answer:

Describe the transformations from the graph of f to the graph of h. Then graph h. (Section 10.1 and Section 10.2)
Question 13.
f(x) = (sqrt) h(x) = -3 (sqrt) + 6
Answer:

Question 15.
The time t (in seconds) it takes a dropped object to fall h feet is given by t = (frac<1> <4>sqrt). (Section 10.1)

a. Use a graphing calculator to graph the function. Describe the domain and range.
b. It takes about 7.4 seconds for a stone dropped from the New River Gorge Bridge in West Virginia to reach the water below. About how high is the bridge above the New River?
Answer:

Question 16.
The radius r of a sphere is given by the function r = (sqrt[3] <4 pi>V>), where V is the volume of the sphere. Spaceship Earth is a spherical structure at Walt Disney World that has an inner radius of about 25 meters. Use a graphing calculator to graph the function. Estimate the volume of Spaceship Earth. (Section 10.2)
Answer:

Question 17.
The graph of square root function g is shown. Compare the average rate of change of g to the average rate of change of h(x) = (sqrt[3] <2>x>)x over the interval x = 0 to x = 3.

Answer:

Lesson 10.3 Solving Radical Equations

Essential Question How can you solve an equation that contains square roots?

EXPLORATION 1

Analyzing a Free-Falling Object
Work with a partner. The table shows the time t (in seconds) that it takes a free-falling object (with no air resistance) to fall d feet.

a. Use the data in the table to sketch the graph of t as a function of d. Use the coordinate plane below.
b. Use your graph to estimate the time it takes the object to fall 240 feet.
c. The relationship between d and t is given by the function t = (sqrt<16>>).
Use this function to check your estimate in part (b).
d. It takes 5 seconds for the object to hit the ground. How far did it fall? Explain your reasoning.

EXPLORATION 2

Solving a Square Root Equation
Work with a partner. The speed s (in feet per second) of the free-falling object in Exploration 1 is given by the functions
s = (sqrt<64d>).
Find the distance the object has fallen when it reaches each speed.
a. s = 8 ft/sec
b. s = 16 ft/sec
c. s = 24 ft/sec

Communicate Your Answer

Question 3.
How can you solve an equation that contains square roots?
Answer:

Question 4.
Use your answer to Question 3 to solve each equation.
a. 5 = (sqrt) + 20
b. 4 = (sqrt)
c. (sqrt) + 2 = 3
d. -3 = -2(sqrt)
Answer:

Monitoring Progress

Solve the equation. Check your solution.
Question 1.
(sqrt) = 6
Answer:

Question 2.
(sqrt) – 7 = 3
Answer:

Question 3.
(sqrt) + 15 = 22
Answer:

Solve the equation. Check your solution.
Question 5.
(sqrt) + 7 = 11
Answer:

Question 6.
15 = 6 + (sqrt<3w-9>)
Answer:

Question 9.
(sqrt [3]< y >) = 4 = 1
Answer:

Question 10.
(sqrt [3]< 3c+7 >) = 10
Answer:

Solve the equation. Check your solution(s).
Question 11.
(sqrt<4-3x>) = x
Answer:

Question 12.
(sqrt<3m>) + 10 = 1
Answer:

Question 13.
p + 1 = (sqrt<7p+15>)
Answer:

Question 14.
What is the length of a pendulum that has a period of 2.5 seconds?
Answer:

Solving Radical Equations 10.3 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Why should you check every solution of a radical equation?
Answer:

Question 2.
WHICH ONE DOESN’T BELONG?
Which equation does not belong with the other three? Explain your reasoning.

Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the equation. Check your solution.
Question 3.
(sqrt) = 9
Answer:

Question 4.
(sqrt) = 4
Answer:

Question 5.
7 = (sqrt) – 5
Answer:

Question 7.
(sqrt) + 12 = 23
Answer:

Question 8.
(sqrt) + 6 = 8
Answer:

Question 9.
4 – (sqrt) = 2
Answer:

Question 10.
-8 = 7 = (sqrt)
Answer:

Question 11.
3(sqrt) – 18 = -3
Answer:

Question 12.
2(sqrt) + 5 = 11
Answer:

In Exercises 13–20, solve the equation. Check your solution.
Question 13.
(sqrt) + 5 = 9
Answer:

Question 15.
2 (sqrt) = 16
Answer:

Question 16.
5(sqrt) = 10
Answer:

Question 17.
-1 = (sqrt<5r+1>) – 7
Answer:

Question 19.
7 + 3(sqrt<3p-9>) = 25
Answer:

Question 20.
19 – 4(sqrt<3c-11>) = 11
Answer:

Question 21.
MODELING WITH MATHEMATICS
The Cave of Swallows is a natural open-air pit cave in the state of San Luis Potosí, Mexico. The 1220-foot- deep cave was a popular destination for BASE jumpers. The function t = (frac<1> <4>sqrt) represents the time t (in seconds) that it takes a BASE jumper to fall d feet. How far does a BASE jumper fall in 3 seconds?

Answer:

Question 22.
MODELING WITH MATHEMATICS
The edge length s of a cube with a surface area of A is given by s = (sqrt>). What is the surface area of a cube with an edge length of 4 inches?

Answer:

In Exercises 23–26, use the graph to solve the equation.
Question 23.

Answer:

Question 24.

Answer:

Question 25.

Answer:

Question 26.

Answer:

In Exercises 27–34, solve the equation. Check your solution. (See Example 3.)
Question 27.
(sqrt<2x-9>) = (sqrt)
Answer:

Question 29.
(sqrt<3g+1>) = (sqrt<7g-19>)
Answer:

Question 31.
(sqrt<2>-2>) = (sqrt)
Answer:

Question 33.
(sqrt<2c+1>) = (sqrt<4c>) = 0
Answer:

MATHEMATICAL CONNECTIONS In Exercises 35 and 36, find the value of x.
Question 35.
Perimeter = 22 cm

Answer:

Question 36.

Answer:

In Exercises 37–44, solve the equation. Check your solution.
Question 37.
(sqrt [3]< x >) = 4
Answer:

Question 38.
(sqrt [3]< y >) = 2
Answer:

Question 39.
6 = 3(sqrt [3]< 8g >)
Answer:

Question 40.
(sqrt [3]< r+19 >) = 3
Answer:

Question 41.
(sqrt [3]< 2x+9 >) = -3
Answer:

Question 43.
(sqrt [3]< y+6 >) = (sqrt [3]< 5y-2 >)
Answer:

In Exercises 45–48, determine which solution, if any, is an extraneous solution.
Question 45.
(sqrt<6x-5>) = x x = 5, x = 1
Answer:

Question 46.
(sqrt<2y+3>) = y y = -1, y = 3
Answer:

Question 47.
(sqrt<12p+16>) = -2p p = -1, p = 4
Answer:

Question 48.
-3g = (sqrt<-18-27>) g = -2, g = -1
Answer:

In Exercises 49–58, solve the equation. Check your solution(s).
Question 49.
y = (sqrt<5y-4>)
Answer:

Question 51.
(sqrt<1-3a>) = 2a
Answer:

Question 52.
2q = (sqrt<10q-6>)
Answer:

Question 53.
9 + (sqrt<5p>) = 4
Answer:

Question 54.
(sqrt<3n>) – 11 = -5
Answer:

Question 55.
(sqrt<2m+2>) – 3 = 1
Answer:

Question 56.
15 + (sqrt<4b-8>) = 13
Answer:

Question 57.
r + 4 = (sqrt<-4r-19>)
Answer:

ERROR ANALYSIS In Exercises 59 and 60, describe and correct the error in solving the equation.
Question 59.

Answer:

Question 60.

Answer:

Question 61.
REASONING
Explain how to use mental math to solve (sqrt<2x>) + 5 = 1.
Answer:

Question 62.
WRITING
Explain how you would solve (sqrt [4]< m+4 >) – (sqrt [4]< 3m >) = 0.
Answer:

Question 63.
MODELING WITH MATHEMATICS
The formula V = (sqrt) relates the voltage V (in volts), power P (in watts), and resistance R (in ohms) of an electrical circuit. The hair dryer shown is on a 120-volt circuit. Is the resistance of the hair dryer half as much as the resistance of the same hair dryer on a 240-volt circuit? Explain your reasoning.

Answer:

Question 64.
MODELING WITH MATHEMATICS
The time t (in seconds) it takes a trapeze artist to swing back and forth is represented by the function t = 2π (sqrt<32>>), where r is the rope length (in feet). It takes the trapeze artist 6 seconds to swing back and forth. Is this rope (frac<3><2>) as long as the rope used when it takes the trapeze artist 4 seconds to swing back and forth? Explain your reasoning.

Answer:

REASONING In Exercises 65–68, determine whether the statement is true or false. If it is false, explain why.
Question 65.
If (sqrt) = b, then ((sqrt)) 2 = b 2 .
Answer:

Question 67.
If a 2 = b 2 , then a = b.
Answer:

Question 68.
If a 2 = (sqrt), then a 4 = ((sqrt)) 2
Answer:

Question 69.
COMPARING METHODS
Consider the equation x + 2 = (sqrt<2x-3>).
a. Solve the equation by graphing. Describe the process.
b. Solve the equation algebraically. Describe the process.
c. Which method do you prefer? Explain your reasoning.
Answer:

Question 70.
HOW DO YOU SEE IT?
The graph shows two radical functions.

a. Write an equation whose solution is the x-coordinate of the point of intersection of the graphs.
b. Use the graph to solve the equation.
Answer:

Question 71.
MATHEMATICAL CONNECTIONS
The slant height s of a cone with a radius of r and a height of h is given by s = (sqrt+h^<2>>). The slant heights of the two cones are equal. Find the radius of each cone.

Answer:

Question 72.
CRITICAL THINKING
How is squaring (sqrt) different from squaring (sqrt) + 2?
Answer:

USING STRUCTURE In Exercises 73–78, solve the equation. Check your solution.
Question 73.
(sqrt) = (sqrt) + (sqrt<5>)
Answer:

Question 75.
(sqrt<5y+9>) + (sqrt<5y>) = 9
Answer:

Question 77.
2(sqrt<1+4h>) – 4(sqrt) – 2 = 0
Answer:

Question 79.
OPEN-ENDED
Write a radical equation that has a solution of x = 5.
Answer:

Question 80.
OPEN-ENDED
Write a radical equation that has x = 3 and x = 4 as solutions.
Answer:

Question 81.
MAKING AN ARGUMENT
Your friend says the equation (sqrt<(2 x+5)^<2>>) = 2x + 5 is always true, because after simplifying the left side of the equation, the result is an equation with infinitely many solutions. Is your friend correct? Explain.
Answer:

Question 82.
THOUGHT PROVOKING
Solve the equation (sqrt [3]< x+1 >) = (sqrt). Show your work and explain your steps.
Answer:

Question 83.
MODELING WITH MATHEMATICS
The frequency f (in cycles per second) of a string of an electric guitar is given by the equation f = (frac<1> <2 ell>sqrt>), where ℓ is the length of the string (in meters), T is the string’s tension (in newtons), and m is the string’s mass per unit length (in kilograms per meter). The high E string of an electric guitar is 0.64 meter long with a mass per unit length of 0.000401 kilogram per meter.

a. How much tension is required to produce a frequency of about 330 cycles per second?
b. Would you need more or less tension to create the same frequency on a string with greater mass per unit length? Explain.
Answer:

Maintaining Mathematical Proficiency

Find the product.(Section 7.2)
Question 84.
(x + 8)(x – 2)
Answer:

Question 85.
(3p – 1)(4p + 5)
Answer:

Question 86.
(s + 2)(s 2 + 3s – 4)
Answer:

Graph the function. Compare the graph to the graph of f(x) = x 2 .(Section 8.1)
Question 87.
r(x) = 3x 2
Answer:

Question 88.
g(x) = (frac<3><4>)x 2
Answer:

Question 89.
h(x) = -5x 2
Answer:

Lesson 10.4 Inverse of a Function

Essential Question How are a function and its inverse related?

EXPLORATION 1

Exploring Inverse Functions
Work with a partner. The functions f and g are inverses of each other. Compare the tables of values of the two functions. How are the functions related?

EXPLORATION 2

Exploring Inverse Functions
Work with a partner.
a. Plot the two sets of points represented by the tables in Exploration 1. Use the coordinate plane below.
b. Connect each set of points with a smooth curve.
c. Describe the relationship between the two graphs.
d. Write an equation for each function.

Communicate Your Answer

Question 3.
How are a function and its inverse related?
Answer:

Question 4.
A table of values for a function f is given. Create a table of values for a function g, the inverse of f.

Answer:

Question 5.
Sketch the graphs of f(x) = x + 4 and its inverse in the same coordinate plane. Then write an equation of the inverse of f. Explain your reasoning.

Answer:

Monitoring Progress

Find the inverse of the relation.
Question 1.
(-3, -4), (-2, 0), (-1, 4), (0, 8), (1, 12), (2, 16), (3, 20)
Answer:

Question 2.

Answer:

Solve y = f(x) for x. Then find the input when the output is 4.
Question 3.
f(x) = x – 6
Answer:

Question 4.
f(x) = (frac<1><2>)x + 3
Answer:

Question 5.
f(x) = 4x 2
Answer:

Find the inverse of the function. Then graph the function and its inverse.
Question 6.
f(x) = 6x
Answer:

Question 7.
f(x) = -x + 5
Answer:

Find the inverse of the function. Then graph the function and its inverse.
Question 9.
f(x) = -x 2 , x ≤ 0
Answer:

Question 10.
f(x) = 4x 2 + 3, x ≥ 0
Answer:

Question 11.
Is the inverse of f(x) = (sqrt<2x-1>) a function? Find the inverse.
Answer:

Inverse of a Function 10.4 Exercises

Vocabulary and Core ConceptCheck

Question 1.
COMPLETE THE SENTENCE
A relation contains the point (-3, 10). The ____________ contains the point (10, -3).
Answer:

Question 2.
DIFFERENT WORDS, SAME QUESTION
Consider the function f represented by the graph. Which is different? Find “both” answers.

Answer:

In Exercises 3–8, find the inverse of the relation.
Question 3.
(1, 0), (3, -8), (4, -3), (7, -5), (9, -1)
Answer:

Question 4.
(2, 1), (4, -3), (6, 7), (8, 1), (10, -4)
Answer:

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

In Exercises 9–14, solve y = f(x) for x. Then find the input when the output is 2.
Question 9.
f(x) = x + 5
Answer:

Question 10.
f(x) = 2x – 3
Answer:

Question 11.
f(x) = (frac<1><4>)x – 1
Answer:

Question 12.
f(x) = (frac<2><3>)x + 4
Answer:

Question 13.
f(x) = 9x 2
Answer:

Question 14.
f(x) = (frac<1><2>)x 2 – 7
Answer:

In Exercises 15 and 16, graph the inverse of the function by reflecting the graph in the line y = x. Describe the domain and range of the inverse.
Question 15.

Answer:

Question 16.

Answer:

In Exercises 17–22, find the inverse of the function. Then graph the function and its inverse.
Question 17.
f(x) = 4x – 1
Answer:

Question 18.
f(x) = -2x + 5
Answer:

Question 19.
f(x) = -3x – 2
Answer:

Question 20.
f(x) = 2x + 3
Answer:

Question 21.
f(x) =(frac<1><3>)x + 8
Answer:

In Exercises 23–28, find the inverse of the function. Then graph the function and its inverse.
Question 23.
f(x) = 4x 2 , x ≥ 0
Answer:

Question 24.
f(x) = (frac<2><25>)x 2 , x ≤ 0
Answer:

Question 25.
f(x) = -x 2 + 10, x ≤ 0
Answer:

Question 26.
f(x) = 2x 2 + 6, x ≥ 0
Answer:

Question 27.
f(x) = (frac<1><9>)x 2 + 2, x ≥ 0
Answer:

Question 28.
f(x) = -4x 2 – 8, x ≤ 0
Answer:

In Exercises 29–32, use the Horizontal Line Test to determine whether the inverse of f is a function.
Question 29.

Answer:

Question 30.

Answer:

Question 31.

Answer:

Question 32.

Answer:

In Exercises 33–42, determine whether the inverse of f is a function. Then find the inverse.
Question 33.
f(x) = (sqrt)
Answer:

Question 34.
f(x) = (sqrt)
Answer:

Question 35.
f(x) = (sqrt<2x-6>)
Answer:

Question 37.
f(x) = 3(sqrt)
Answer:

Question 39.
f(x) = –(sqrt<3x+5>) – 2
Answer:

Question 40.
f(x) = 2(sqrt) + 6
Answer:

Question 41.
f(x) = 2x 2
Answer:

Question 43.
ERROR ANALYSIS
Describe and correct the error in finding the inverse of the function f(x) = – 3x + 5.

Answer:

Question 44.
ERROR ANALYSIS
Describe and correct the error in finding and graphing the inverse of the function f(x) = (sqrt).

Answer:

Question 45.
MODELING WITH MATHEMATICS
The euro is the unit of currency for the European Union. On a certain day, the number E of euros that could be obtained for D U.S. dollars was represented by the formula shown.
E = 0.74683D
Solve the formula for D. Then find the number of U.S. dollars that could be obtained for 250 euros on that day.
Answer:

Question 46.
MODELING WITH MATHEMATICS
A crow is flying at a height of 50 feet when it drops a walnut to break it open. The height h (in feet) of the walnut above ground can be modeled by h = -16t 2 + 50, where t is the time (in seconds) since the crow dropped the walnut. Solve the equation for t. After how many seconds will the walnut be 15 feet above the ground?

Answer:

MATHEMATICAL CONNECTIONS In Exercises 47 and 48, s is the side length of an equilateral triangle. Solve the formula for s. Then evaluate the new formula for the given value.

Answer:
47.

In Exercises 49–54, find the inverse of the function. Then graph the function and its inverse.
Question 49.
f(x) = 2x 3
Answer:

Question 50.
f(x) = x 3 – 4
Answer:

Question 51.
f(x) = (x – 5) 3
Answer:

Question 52.
f(x) = 8(x + 2) 3
Answer:

Question 53.
f(x) = 4(sqrt [3]< x >)
Answer:

Question 55.
MAKING AN ARGUMENT
Your friend says that the inverse of the function f(x) = 3 is a function because all linear functions pass the Horizontal Line Test. Is your friend correct? Explain.
Answer:

Question 56.
HOW DO YOU SEE IT?
Pair the graph of each function with the graph of its inverse.

Answer:

Question 57.
WRITING
Describe changes you could make to the function f(x) = x 2 – 5 so that its inverse is a function. Describe the domain and range of the new function and its inverse.
Answer:

Question 58.
CRITICAL THINKING
Can an even function with at least two values in its domain have an inverse that is a function? Explain.
Answer:

Question 59.
OPEN-ENDED
Write a function such that the graph of its inverse is a line with a slope of 4.
Answer:

Question 60.
CRITICAL THINKING
Consider the function g(x) = -x.
a. Graph g(x) = -x and explain why it is its own inverse.
b. Graph other linear functions that are their own inverses. Write equations of the lines you graph.
c. Use your results from part (b) to write a general equation that describes the family of linear functions that are their own inverses.
Answer:

Question 61.
REASONING
Show that the inverse of any linear function f(x) = mx + b, where m ≠ 0, is also a linear function. Write the slope and y-intercept of the graph of the inverse in terms of m and b.
Answer:

Question 62.
THOUGHT PROVOKING
The graphs of f(x) = x 3 – 3x and its inverse are shown. Find the greatest interval -a ≤ x ≤ a for which the inverse of f is a function. Write an equation of the inverse function.

Answer:

Question 63.
REASONING
Is the inverse of f(x) = 2|x + 1|a function? Are there any values of a, h, and k for which the inverse of f(x) = a |x – h| + k is a function? Explain your reasoning.
Answer:

Maintaining Mathematical Proficiency

Find the sum or difference.(Section 7.1)
Question 64.
(2x – 9) – (6x + 5)
Answer:

Question 65.
(8y + 1) + (-y – 12)
Answer:

Question 66.
(t 2 – 4t – 4) + (7t 2 + 12t + 3)
Answer:

Question 67.
(-3d 2 + 10d – 8) – (7d 2 – d – 6)
Answer:

Graph the function. Compare the graph to the graph of f(x) = x 2 . (Section 8.2)
Question 68.
g(x) = x 2 + 6
Answer:

Question 69.
h(x) = -x 2 – 2
Answer:

Question 70.
p(x) = -4x 2 + 5
Answer:

Question 71.
q(x) = (frac<1><3>)x 2 – 1
Answer:

Radical Functions and Equations Performance Task: Medication and the Mosteller Formula

10.3–10.4What Did YouLearn?

Core Vocabulary
radical equation, p. 560
inverse relation, p. 568
inverse function, p. 569

Core Concepts
Lesson 10.3
Squaring Each Side of an Equation, p. 560
Identifying Extraneous Solutions, p. 562

Lesson 10.4
Inverse Relation, p. 568
Finding Inverses of Functions Algebraically, p. 570
Finding Inverses of Nonlinear Functions, p. 570
Horizontal Line Test, p. 571

Mathematical Practices
Question 1.
Could you also solve Exercises 37–44 on page 565 by graphing? Explain.
Answer:

Question 2.
What external resources could you use to check the reasonableness of your answer in Exercise 45 on page 573?
Answer:

Performance Task: Medication and the Mosteller Formula

When taking medication, it is critical to take the correct dosage. For children in particular, body surface area (BSA) is a key component in calculating that dosage. The Mosteller Formula is commonly used to approximate body surface area. How will you use this formula to calculate BSA for the optimum dosage?
To explore the answers to this question and more, go to

Radical Functions and Equations Chapter Review

10.1 Graphing Square Root Functions (pp. 543–550)

Graph the function. Describe the domain and range. Compare the graph to the graph of f(x) = (sqrt).
Question 1.
g(x) = (sqrt) + 7
Answer:

Question 4.
Let g(x) = (frac<1> <4>sqrt) + 2. Describe the transformations from the graph of f(x) = (sqrt) to the graph of g. Then graph g.
Answer:

10.2 Graphing Cube Root Functions (pp. 551–556)

Graph the function. Compare the graph to the graph of f(x) = (sqrt [3]< x >).
Question 5.
g(x) = (sqrt [3]< x >) + 4
Answer:

Question 6.
h(x) = -8(sqrt [3]< x >)
Answer:

Question 8.
Let g(x) = -3(sqrt [3]< x+2 >) – 1. Describe the transformations from the graph of f(x) = (sqrt [3]< x >) to the graph of g. Then graph g.
Answer:

Question 9.
The graph of cube root function r is shown. Compare the average rate of change of r to the average rate of change of p(x) = (sqrt[3] <2>x>) over the interval x = 0 to x = 8.
Answer:

10.3 Solving Radical Equations (pp. 559-566)

Solve the equation. Check your solution(s).
Question 10.
8 + (sqrt) = 18
Answer:

Question 11.
(sqrt [3]< x-1 >) = 3
Answer:

Question 13.
x = (sqrt<3x+4>)
Answer:

Question 14.
8(sqrt) + 34 = 58
Answer:

Question 15.
(sqrt<5x>) + 6 = 5
Answer:

Question 16.
The radius r of a cylinder is represented by the function r = (sqrt>), where V is the volume and h is the height of the cylinder. What is the volume of the cylindrical can?

Answer:

10.4 Inverse of a Function (pp. 567–574)

Find the inverse of the relation.
Question 17.
(1, -10), (3, -4), (5, 4), (7, 14), (9, 26)
Answer:

Question 18.

Answer:

Find the inverse of the function. Then graph the function and its inverse.
Question 19.
f(x) = -5x + 10
Answer:

Question 20.
f(x) = 3x 2 – 1, x ≥ 0
Answer:

Question 22.
Consider the function f(x) = x 2 + 4. Use the Horizontal Line Test to determine whether the inverse of f is a function.
Answer:

Question 23.
In bowling, a handicap is an adjustment to a bowler’s score to even out differences in ability levels. In a particular league, you can find a bowler’s handicap h by using the formula h = 0.8(210 – a), where a is the bowler’s average. Solve the formula for a. Then find a bowler’s average when the bowler’s handicap is 28.

Answer:

Radical Functions and Equations Chapter Test

Find the inverse of the function.
Question 1.
f(x) = 5x – 8 2.
Answer:

Question 3.
f(x) = –(frac<1><3>)x 2 + 4, x ≥ 0
Answer:

Graph the function f. Describe the domain and range. Compare the graph of f to the graph of g.
Question 4.
f(x) = –(sqrt) g(x) = (sqrt)
Answer:

Question 5.
f(x) = (sqrt) + 2 g(x) = (sqrt)
Answer:

Question 6.
f(x) = (sqrt [3]< x >) – 5 g(x) = (sqrt [3]< x >)
Answer:

Question 7.
f(x) = -2(sqrt [3]< x+1 >) g(x) = (sqrt [3]< x >)
Answer:

Solve the equation. Check your solution(s).
Question 8.
9 – (sqrt) = 3
Answer:

Question 11.
x + 5 = (sqrt<7x+53>)
Answer:

Question 12.
When solving the equation x – 5 = (sqrt), you obtain x = 2 and x = 8. Explain why at least one of these solutions must be extraneous.
Answer:

Describe the transformations from the graph of f(x) = (sqrt [3]< x >) to the graph of the given function. Then graph the given function.
Question 13.
h(x) = 4(sqrt [3]< x-1 >) + 5
Answer:

Question 15.
The velocity v (in meters per second) of a roller coaster at the bottom of a hill is given by v = (sqrt<19.6h>), where h is the height (in meters) of the hill. (a) Use a graphing calculator to graph the function. Describe the domain and range. (b) How tall must the hill be for the velocity of the roller coaster at the bottom of the hill to be at least 28 meters per second? (c) What happens to the average rate of change of the velocity as the height of the hill increases?
Answer:

Question 16.
The speed s (in meters per second) of sound through air is given by s = 2(sqrt), where T is the temperature (in degrees Celsius).

a. What is the temperature when the speed of sound through air is 340 meters per second?
b. How long does it take you to hear the wolf howl when the temperature is -17°C?
Answer:

Question 17.
How can you restrict the domain of the function f(x) = (x – 3) 2 so that the inverse of f is a function?
Answer:

Question 18.
Write a radical function that has a domain of all real numbers less than or equal to 0 and a range of all real numbers greater than or equal to 9.
Answer:

Radical Functions and Equations Cumulative Assessment

Question 1.
Fill in the function so that it is represented by the graph.

Answer:

Question 2.
Consider the equation y = mx + b. Fill in values for m and b so that each statement is true.
a. When m = ______ and b = ______, the graph of the equation passes through the point (-1, 4).
b. When m = ______ and b = ______, the graph of the equation has a positive slope and passes through the point (-2, -5).
c. When m = ______ and b = ______, the graph of the equation is perpendicular to the graph of y = 4x – 3 and passes through the point (1, 6).
Answer:

Question 3.
Which graph represents the inverse of the function f(x) = 2x + 4?

Answer:

Question 4.
Consider the equation x = (sqrt). Student A claims this equation has one real solution. Student B claims this equation has two real solutions. Use the numbers to answer parts (a)–(c).

a. Choose values for a and b to create an equation that supports Student A’s claim.
b. Choose values for a and b to create an equation that supports Student B’s claim.
c. Choose values for a and b to create an equation that does not support either student’s claim.
Answer:

Question 5.
Which equation represents the nth term of the sequence 3, 12, 48, 192, . . .?
A. an = 3(4) n-1
B. an = 3(9) n-1
C. an = 9n – 6
D. an = 9n + 3
Answer:

Question 6.
Consider the function f(x) = (frac<1> <2>sqrt[3]). The graph represents function g. Select all the statements that are true.

Answer:

Question 7.
Place each function into one of the three categories.

Answer:

Question 8.
You are making a tabletop with a tiled center and a uniform mosaic border.
a. Write the polynomial in standard form that represents the perimeter of the tabletop.
b. Write the polynomial in standard form that represents the area of the tabletop.
c. The perimeter of the tabletop is less than 80 inches, and the area of tabletop is at least 252 square inches. Select all the possible values of x.

Answer:


Math Word Problems and Solutions - Distance, Speed, Time

Problem 1 A salesman sold twice as much pears in the afternoon than in the morning. If he sold 360 kilograms of pears that day, how many kilograms did he sell in the morning and how many in the afternoon?
Click to see solution

Problem 2 Mary, Peter, and Lucy were picking chestnuts. Mary picked twice as much chestnuts than Peter. Lucy picked 2 kg more than Peter. Together the three of them picked 26 kg of chestnuts. How many kilograms did each of them pick?
Click to see solution

Problem 3
Sophia finished $frac<2><3>$ of a book. She calculated that she finished 90 more pages than she has yet to read. How long is her book?
Click to see solution

Problem 4
A farming field can be ploughed by 6 tractors in 4 days. When 6 tractors work together, each of them ploughs 120 hectares a day. If two of the tractors were moved to another field, then the remaining 4 tractors could plough the same field in 5 days. How many hectares a day would one tractor plough then?
Click to see solution

Problem 5
A student chose a number, multiplied it by 2, then subtracted 138 from the result and got 102. What was the number he chose?
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Problem 6
I chose a number and divide it by 5. Then I subtracted 154 from the result and got 6. What was the number I chose?
Click to see solution

Problem 7
The distance between two towns is 380 km. At the same moment, a passenger car and a truck start moving towards each other from different towns. They meet 4 hours later. If the car drives 5 km/hr faster than the truck, what are their speeds?
Click to see solution

Problem 8
One side of a rectangle is 3 cm shorter than the other side. If we increase the length of each side by 1 cm, then the area of the rectangle will increase by 18 cm 2 . Find the lengths of all sides.
Click to see solution

Problem 9
The first year, two cows produced 8100 litres of milk. The second year their production increased by 15% and 10% respectively, and the total amount of milk increased to 9100 litres a year. How many litres were milked from each cow each year?
Click to see solution

Problem 10
The distance between stations A and B is 148 km. An express train left station A towards station B with the speed of 80 km/hr. At the same time, a freight train left station B towards station A with the speed of 36 km/hr. They met at station C at 12 pm, and by that time the express train stopped at at intermediate station for 10 min and the freight train stopped for 5 min. Find:
a) The distance between stations C and B.
b) The time when the freight train left station B.
Click to see solution

Problem 11
Susan drives from city A to city B. After two hours of driving she noticed that she covered 80 km and calculated that, if she continued driving at the same speed, she would end up been 15 minutes late. So she increased her speed by 10 km/hr and she arrived at city B 36 minutes earlier than she planned.
Find the distance between cities A and B.
Click to see solution

Problem 12
To deliver an order on time, a company has to make 25 parts a day. After making 25 parts per day for 3 days, the company started to produce 5 more parts per day, and by the last day of work 100 more parts than planned were produced. Find how many parts the company made and how many days this took.
Click to see solution

Problem 13
There are 24 students in a seventh grade class. They decided to plant birches and roses at the school's backyard. While each girl planted 3 roses, every three boys planted 1 birch. By the end of the day they planted $24$ plants. How many birches and roses were planted?
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Problem 14
A car left town A towards town B driving at a speed of V = 32 km/hr. After 3 hours on the road the driver stopped for 15 min in town C. Because of a closed road he had to change his route, making the trip 28 km longer. He increased his speed to V = 40 km/hr but still he was 30 min late. Find:
a) The distance the car has covered.
b) The time that took it to get from C to B.
Click to see solution

Solution:
From the statement of the problem we don't know if the 15 min stop in town C was planned or it was unexpected. So we have to consider both cases.

A
The stop was planned. Let us consider only the trip from C to B, and let $x$ be the number of hours the driver spent on this trip.
Then the distance from C to B is $S = 40cdot x$ km. If the driver could use the initial route, it would take him $x - frac<30> <60>= x - frac<1><2>$ hours to drive from C to B. The distance from C to B according to the initially itinerary was $(x - frac<1><2>)cdot 32$ km, and this distance is $28$ km shorter than $40cdot x$ km. Then we have the equation
$(x - 1/2)cdot 32 + 28 = 40x$
$32x -16 +28 = 40x$
$-8x = -12$
$8x = 12$
$x = frac<12><8>$
$x = 1 frac<4> <8>= 1 frac<1> <2>= 1 frac<30> <60>=$ 1 hr 30 min.
So, the car covered the distance between C and B in 1 hour and 30 min.
The distance from A to B is $3cdot 32 + frac<12><8>cdot 40 = 96 + 60 = 156$ km.

B
Suppose it took $x$ hours for him to get from C to B. Then the distance is $S = 40cdot x$ km.
The driver did not plan the stop at C. Let we accept that he stopped because he had to change the route.
It took $x - frac<30> <60>+ frac<15> <60>= x - frac<15> <60>= x - frac<1><4>$ h to drive from C to B. The distance from C to B is $32(x - frac<1><4>)$ km, which is $28$ km shorter than $40cdot x$, i.e.
$32(x - frac<1><4>) + 28 = 40x$
$32x - 8 +28 = 40x$
$20= 8x$
$x = frac<20> <8>= frac<5> <2>= 2 ext


30 ext.$
The distance covered equals $ 40 imes 2.5 = 100 km$.

Problem 15
If a farmer wants to plough a farm field on time, he must plough 120 hectares a day. For technical reasons he ploughed only 85 hectares a day, hence he had to plough 2 more days than he planned and he still has 40 hectares left. What is the area of the farm field and how many days the farmer planned to work initially?
Click to see solution

Problem 16
A woodworker normally makes a certain number of parts in 24 days. But he was able to increase his productivity by 5 parts per day, and so he not only finished the job in only 22 days but also he made 80 extra parts. How many parts does the woodworker normally makes per day and how many pieces does he make in 24 days?
Click to see solution

Problem 17
A biker covered half the distance between two towns in 2 hr 30 min. After that he increased his speed by 2 km/hr. He covered the second half of the distance in 2 hr 20 min. Find the distance between the two towns and the initial speed of the biker.
Click to see solution

Problem 18
A train covered half of the distance between stations A and B at the speed of 48 km/hr, but then it had to stop for 15 min. To make up for the delay, it increased its speed by $frac<5><3>$ m/sec and it arrived to station B on time. Find the distance between the two stations and the speed of the train after the stop.
Click to see solution

Problem 19
Elizabeth can get a certain job done in 15 days, and Tony can finish only 75% of that job within the same time. Tony worked alone for several days and then Elizabeth joined him, so they finished the rest of the job in 6 days, working together.
For how many days have each of them worked and what percentage of the job have each of them completed?
Click to see solution

Problem 20
A farmer planned to plough a field by doing 120 hectares a day. After two days of work he increased his daily productivity by 25% and he finished the job two days ahead of schedule.
a) What is the area of the field?
b) In how many days did the farmer get the job done?
c) In how many days did the farmer plan to get the job done?
Click to see solution

Problem 21
To mow a grass field a team of mowers planned to cover 15 hectares a day. After 4 working days they increased the daily productivity by $33 imes frac<1><3>\%$, and finished the work 1 day earlier than it was planned.
A) What is the area of the grass field?
B) How many days did it take to mow the whole field?
C) How many days were scheduled initially for this job?
Hint : See problem 20 and solve by yourself.
Answer: A) 120 hectares B) 7 days C) 8 days.

Problem 22
A train travels from station A to station B. If the train leaves station A and makes 75 km/hr, it arrives at station B 48 minutes ahead of scheduled. If it made 50 km/hr, then by the scheduled time of arrival it would still have 40 km more to go to station B. Find:
A) The distance between the two stations
B) The time it takes the train to travel from A to B according to the schedule
C) The speed of the train when it's on schedule.
Click to see solution

Problem 23
The distance between towns A and B is 300 km. One train departs from town A and another train departs from town B, both leaving at the same moment of time and heading towards each other. We know that one of them is 10 km/hr faster than the other. Find the speeds of both trains if 2 hours after their departure the distance between them is 40 km.
Click to see solution

Problem 24
A bus travels from town A to town B. If the bus's speed is 50 km/hr, it will arrive in town B 42 min later than scheduled. If the bus increases its speed by $frac<50><9>$ m/sec, it will arrive in town B 30 min earlier than scheduled. Find:
A) The distance between the two towns
B) The bus's scheduled time of arrival in B
C) The speed of the bus when it's on schedule.
Click to see solution


Solving radical equations: Do non-real extraneous solutions exist?

Is it correct to say that in this example, the two complex values for x are in fact "non-real extraneous solutions"? Relevant Equations: Original equation: x+4 = √(x+10) .
(See work below for further equations.)

Assume you have a radial equation (eg. x+4 = √(x+10) ) that you want to solve for "x".

To solve:
egin (x+4 & = sqrt[2] )^2 onumber
(x+4)^2 &= |x+10| onumber end
For my question, we are only going to consider the case where x+10 < 0:
egin x^2+8x+16 &= -(x+10) onumber
x^2+8x+16 &= -x-10 onumber
0 &= x^2+9x+26 onumber end

Using the quadratic formula, we get:

After checking if LS = RS for the original equation for each complex answer, it seems that neither is a solution because ## LS eq RS ##.

Therefore, what I'd like to know is: In this example, is it correct to say that that the two complex values for x are in fact "non-real extraneous solutions"? If no, then please explain.

(I would also appreciate if anyone has an links to articles about extraneous non-real solutions to radical equations. Most of the articles online that I can easily find basically only talk about extraneous solutions in the context of "real number" answers. I'm not really sure if it makes any mathematical sense to about extraneous complex solutions--I don't know enough about the subject. To be clear, I already know that "extraneous solutions" are invalid solutions to the starting equation I just don't know if the term "extraneous solutions" can be also applied to non-real solutions that happen to be invalid solutions. In other words, I'm having trouble visualizing what a complex extraneous solution would look like graphically--which makes me wonder if such a thing even exists. )


1.4: Radical Equations - Mathematics

Roots and Radicals deserve their own chapter and homework because they occur frequently in applications.

Let be a natural number , and let be a real number . The -th root of is a number that satisfies The number is denoted by

For example, since , and since .

The symbol is called the radical symbol, and an expression involving it is called a radical (expression) .

If then is the square root of and the number is usually omitted. For example,

If , then is the cube root of . For example, the cube root of is , and that of is .

If is even and is positive then there are two -th roots of , each being the negative of the other. For example, since there are two square roots of . In that case by convention the symbol means the positive -th root of , and it is called the principal ( -th) root of .

If is negative and is odd then there is just one -th root, and it is negative also. For example,

At this stage we do not know of an -th root if is even and is negative. This leads to the subject of complex numbers which we will take up later in the course.

Radicals are just special cases of powers, and you can simplify much of your thinking by keeping this fact in mind:

It follows immediately from that observation and the properties of powers that

Solving Radical Equations

An equation involving radicals is called a radical equation (naturally). To solve it you simply apply our general principle:

To solve an equation figure out what bothers you and then do the same thing on both sides of the equation to get rid of it.

To get rid of a radical you take it to a power that will change the rational exponent to a natural number. This will work if the radical is on one side of the equation by itself.

Let's look at a few simple examples:

Here is a slightly more complicated problem:

Our last example shows how to get rid of more than one radical:

To get rid of the square roots we isolate them and square one at a time:

In each case, we check our answer by substituting it in the original equation. For example, in the last equation we obtain:

Later in the course we will consider more complicated cases of radical equations.

Numerical Values

The radicals in the above examples were all natural numbers. This is due only to a judicious choice of examples. Frequently the roots occurring in applications are irrational numbers with decimal expansions that never repeat or terminate. The following table lists approximations of a few specific radicals.


Simplifying Radicals

Sometimes radical expressions can be simplified. The simplest case is when the radicand is a perfect power, meaning that it’s equal to the nth power of a whole number.

For example the perfect squares are: 1, 4, 9, 16, 25, 36, etc., because 1 = 1 2 , 4 = 2 2 , 9 = 3 2 , 16 = 4 2 , 25 = 5 2 , 36 = 6 2 , and so on. Therefore, we have &radic 1 = 1, &radic 4 = 2, &radic 9 = 3, etc.

Perfect cubes include: 1, 8, 27, 64, etc. So, , and so on.

Variables with exponents also count as perfect powers if the exponent is a multiple of the index. For instance, x 2 is a perfect square. But so are y 4 and w 18 .

Working with Perfect Powers

The key concept is that an nth root of a perfect nth power will completely simplify to a non-radical expression.

For example, .

The square root of 81 is 9, while the square root of x 6 is (x 6 ) 1/2 = x 3 .

Pulling Out Perfect Power Factors

If the radicand is not a perfect power, then you still might be able to factor it into part that is a perfect power. Then you can simplify that part of it and leave the rest within the radical.

Let me show you what I mean.

Suppose you want to take the cube root of 24x 4 . Well 24 is not a perfect cube… but one of its factors, 8, definitely is! Similarly, x 4 is not a perfect cube, but we can factor it as x 3 ·x, and x 3 is a perfect cube.

Notice how the cube root of 8x 3 simplifies to 2x, which we then write outside of the radical expression (in green). The leftover 3x cannot simplify and must remain within the radical.

Rationalizing the Denominator

Finally, we have to discuss another method of simplifying radicals called rationalizing the denominator. This is a technique for rewriting a radical expression in which the radical shows up on the bottom of a fraction (denominator).

There are two basic cases to worry about. Also, for pretty much every problem in your Algebra class, you only have to do this for square roots. (Rationalizing roots with higher index can get a lot more complicated.)

  • The denominator is a single term involving a radical. In this case, multiply both top and bottom of the fraction by the radical expression.
  • The demoninator has two terms, like a + &radic b . Here, you must multiply both the top and bottom of the fraction by the conjugate, a – &radic b . And if you had a – &radic b to begin with, its conjugate would be a + &radic b . (Just flip the sign of the radical.)

Here’s an example showing the technique in practice.


Tegmark's MUH is: Our external physical reality is a mathematical structure. [3] That is, the physical universe is not merely described by mathematics, but is mathematics (specifically, a mathematical structure). Mathematical existence equals physical existence, and all structures that exist mathematically exist physically as well. Observers, including humans, are "self-aware substructures (SASs)". In any mathematical structure complex enough to contain such substructures, they "will subjectively perceive themselves as existing in a physically 'real' world". [4]

The theory can be considered a form of Pythagoreanism or Platonism in that it proposes the existence of mathematical entities a form of mathematical monism in that it denies that anything exists except mathematical objects and a formal expression of ontic structural realism.

Tegmark claims that the hypothesis has no free parameters and is not observationally ruled out. Thus, he reasons, it is preferred over other theories-of-everything by Occam's Razor. Tegmark also considers augmenting the MUH with a second assumption, the computable universe hypothesis (CUH), which says that the mathematical structure that is our external physical reality is defined by computable functions. [5]

The MUH is related to Tegmark's categorization of four levels of the multiverse. [6] This categorization posits a nested hierarchy of increasing diversity, with worlds corresponding to different sets of initial conditions (level 1), physical constants (level 2), quantum branches (level 3), and altogether different equations or mathematical structures (level 4).

Andreas Albrecht of Imperial College in London, called it a "provocative" solution to one of the central problems facing physics. Although he "wouldn't dare" go so far as to say he believes it, he noted that "it's actually quite difficult to construct a theory where everything we see is all there is". [7]

Definition of the ensemble Edit

Jürgen Schmidhuber [8] argues that "Although Tegmark suggests that '. all mathematical structures are a priori given equal statistical weight,' there is no way of assigning equal non-vanishing probability to all (infinitely many) mathematical structures." Schmidhuber puts forward a more restricted ensemble which admits only universe representations describable by constructive mathematics, that is, computer programs e.g., the Global Digital Mathematics Library and Digital Library of Mathematical Functions, linked open data representations of formalized fundamental theorems intended to serve as building blocks for additional mathematical results. He explicitly includes universe representations describable by non-halting programs whose output bits converge after finite time, although the convergence time itself may not be predictable by a halting program, due to the undecidability of the halting problem. [8] [9]

In response, Tegmark notes [3] [ citation needed ] (sec. V.E) that a constructive mathematics formalized measure of free parameter variations of physical dimensions, constants, and laws over all universes has not yet been constructed for the string theory landscape either, so this should not be regarded as a "show-stopper".

Consistency with Gödel's theorem Edit

It has also been suggested that the MUH is inconsistent with Gödel's incompleteness theorem. In a three-way debate between Tegmark and fellow physicists Piet Hut and Mark Alford, [10] the "secularist" (Alford) states that "the methods allowed by formalists cannot prove all the theorems in a sufficiently powerful system. The idea that math is 'out there' is incompatible with the idea that it consists of formal systems."

Tegmark's response in [10] (sec VI.A.1) is to offer a new hypothesis "that only Gödel-complete (fully decidable) mathematical structures have physical existence. This drastically shrinks the Level IV multiverse, essentially placing an upper limit on complexity, and may have the attractive side effect of explaining the relative simplicity of our universe." Tegmark goes on to note that although conventional theories in physics are Gödel-undecidable, the actual mathematical structure describing our world could still be Gödel-complete, and "could in principle contain observers capable of thinking about Gödel-incomplete mathematics, just as finite-state digital computers can prove certain theorems about Gödel-incomplete formal systems like Peano arithmetic." In [3] (sec. VII) he gives a more detailed response, proposing as an alternative to MUH the more restricted "Computable Universe Hypothesis" (CUH) which only includes mathematical structures that are simple enough that Gödel's theorem does not require them to contain any undecidable or uncomputable theorems. Tegmark admits that this approach faces "serious challenges", including (a) it excludes much of the mathematical landscape (b) the measure on the space of allowed theories may itself be uncomputable and (c) "virtually all historically successful theories of physics violate the CUH".

Observability Edit

Stoeger, Ellis, and Kircher [11] (sec. 7) note that in a true multiverse theory, "the universes are then completely disjoint and nothing that happens in any one of them is causally linked to what happens in any other one. This lack of any causal connection in such multiverses really places them beyond any scientific support". Ellis [12] (p29) specifically criticizes the MUH, stating that an infinite ensemble of completely disconnected universes is "completely untestable, despite hopeful remarks sometimes made, see, e.g., Tegmark (1998)." Tegmark maintains that MUH is testable, stating that it predicts (a) that "physics research will uncover mathematical regularities in nature", and (b) by assuming that we occupy a typical member of the multiverse of mathematical structures, one could "start testing multiverse predictions by assessing how typical our universe is" ( [3] sec. VIII.C).

Plausibility of radical Platonism Edit

The MUH is based on the radical Platonist view that math is an external reality ( [3] sec V.C). However, Jannes [13] argues that "mathematics is at least in part a human construction", on the basis that if it is an external reality, then it should be found in some other animals as well: "Tegmark argues that, if we want to give a complete description of reality, then we will need a language independent of us humans, understandable for non-human sentient entities, such as aliens and future supercomputers". Brian Greene ( [14] p. 299) argues similarly: "The deepest description of the universe should not require concepts whose meaning relies on human experience or interpretation. Reality transcends our existence and so shouldn't, in any fundamental way, depend on ideas of our making."

However, there are many non-human entities, plenty of which are intelligent, and many of which can apprehend, memorise, compare and even approximately add numerical quantities. Several animals have also passed the mirror test of self-consciousness. But a few surprising examples of mathematical abstraction notwithstanding (for example, chimpanzees can be trained to carry out symbolic addition with digits, or the report of a parrot understanding a “zero-like concept”), all examples of animal intelligence with respect to mathematics are limited to basic counting abilities. He adds, "non-human intelligent beings should exist that understand the language of advanced mathematics. However, none of the non-human intelligent beings that we know of confirm the status of (advanced) mathematics as an objective language." In the paper "On Math, Matter and Mind" [10] the secularist viewpoint examined argues (sec. VI.A) that math is evolving over time, there is "no reason to think it is converging to a definite structure, with fixed questions and established ways to address them", and also that "The Radical Platonist position is just another metaphysical theory like solipsism. In the end the metaphysics just demands that we use a different language for saying what we already knew." Tegmark responds (sec VI.A.1) that "The notion of a mathematical structure is rigorously defined in any book on Model Theory", and that non-human mathematics would only differ from our own "because we are uncovering a different part of what is in fact a consistent and unified picture, so math is converging in this sense." In his 2014 book on the MUH, Tegmark argues that the resolution is not that we invent the language of mathematics, but that we discover the structure of mathematics.

Coexistence of all mathematical structures Edit

Don Page has argued [15] (sec 4) that "At the ultimate level, there can be only one world and, if mathematical structures are broad enough to include all possible worlds or at least our own, there must be one unique mathematical structure that describes ultimate reality. So I think it is logical nonsense to talk of Level 4 in the sense of the co-existence of all mathematical structures." This means there can only be one mathematical corpus. Tegmark responds ( [3] sec. V.E) that "this is less inconsistent with Level IV than it may sound, since many mathematical structures decompose into unrelated substructures, and separate ones can be unified."

Consistency with our "simple universe" Edit

Alexander Vilenkin comments [16] (Ch. 19, p. 203) that "the number of mathematical structures increases with increasing complexity, suggesting that 'typical' structures should be horrendously large and cumbersome. This seems to be in conflict with the beauty and simplicity of the theories describing our world". He goes on to note (footnote 8, p. 222) that Tegmark's solution to this problem, the assigning of lower "weights" to the more complex structures ( [6] [ citation needed ] sec. V.B) seems arbitrary ("Who determines the weights?") and may not be logically consistent ("It seems to introduce an additional mathematical structure, but all of them are supposed to be already included in the set").

Occam's razor Edit

Tegmark has been criticized as misunderstanding the nature and application of Occam's razor Massimo Pigliucci reminds that "Occam's razor is just a useful heuristic, it should never be used as the final arbiter to decide which theory is to be favored". [17]


1.4: Radical Equations - Mathematics

A radical equation Any equation that contains one or more radicals with a variable in the radicand. is any equation that contains one or more radicals with a variable in the radicand. Following are some examples of radical equations, all of which will be solved in this section:

We begin with the squaring property of equality Given real numbers a and b, where a = b , then a 2 = b 2 . given real numbers a and b, we have the following:

In other words, equality is retained if we square both sides of an equation.

The converse, on the other hand, is not necessarily true:

This is important because we will use this property to solve radical equations. Consider a very simple radical equation that can be solved by inspection:

Here we can see that x = 9 is a solution. To solve this equation algebraically, make use of the squaring property of equality and the fact that ( a ) 2 = a 2 = a when a is positive. Eliminate the square root by squaring both sides of the equation as follows:

As a check, we can see that 9 = 3 as expected. Because the converse of the squaring property of equality is not necessarily true, solutions to the squared equation may not be solutions to the original. Hence squaring both sides of an equation introduces the possibility of extraneous solutions A solution that does not solve the original equation. , or solutions that do not solve the original equation. For this reason, we must check the answers that result from squaring both sides of an equation.

Example 1: Solve: x − 1 = 5 .

Solution: We can eliminate the square root by applying the squaring property of equality.

Answer: The solution is 26.

Example 2: Solve: 5 − 4 x = x .

Solution: Begin by squaring both sides of the equation.

You are left with a quadratic equation that can be solved by factoring.

Since you squared both sides, you must check your solutions.

After checking, you can see that x = − 5 was extraneous it did not solve the original radical equation. Disregard that answer. This leaves x = 1 as the only solution.

Answer: The solution is x = 1 .

In the previous two examples, notice that the radical is isolated on one side of the equation. Typically, this is not the case. The steps for solving radical equations involving square roots are outlined in the following example.

Example 3: Solve: 2 x − 5 + 4 = x .

Step 1: Isolate the square root. Begin by subtracting 4 from both sides of the equation.

Step 2: Square both sides. Squaring both sides eliminates the square root.

Step 3: Solve the resulting equation. Here you are left with a quadratic equation that can be solved by factoring.

Step 4: Check the solutions in the original equation. Squaring both sides introduces the possibility of extraneous solutions hence the check is required.

After checking, we can see that x = 3 is an extraneous root it does not solve the original radical equation. This leaves x = 7 as the only solution.

Answer: The solution is x = 7 .

Example 4: Solve: 3 x + 1 − 2 x = 0 .

Solution: Begin by isolating the term with the radical.

Despite the fact that the term on the left side has a coefficient, it is still considered isolated. Recall that terms are separated by addition or subtraction operators.

Solve the resulting quadratic equation.

Since we squared both sides, we must check our solutions.

After checking, we can see that x = − 3 4 was extraneous.

Sometimes both of the possible solutions are extraneous.

Example 5: Solve: 4 − 11 x − x + 2 = 0 .

Solution: Begin by isolating the radical.

Since we squared both sides, we must check our solutions.

Since both possible solutions are extraneous, the equation has no solution.

The squaring property of equality extends to any positive integer power n. Given real numbers a and b, we have the following:

This is often referred to as the power property of equality Given any positive integer n and real numbers a and b, where a = b , then a n = b n . . Use this property, along with the fact that ( a n ) n = a n n = a , when a is positive, to solve radical equations with indices greater than 2.

Example 6: Solve: x 2 + 4 3 − 2 = 0 .

Solution: Isolate the radical and then cube both sides of the equation.

Answer: The solutions are −2 and 2.

Try this! Solve: 2 x − 1 + 2 = x .

Answer: x = 5 ( x = 1 is extraneous)

Video Solution

It may be the case that the equation has two radical expressions.

Example 7: Solve: 3 x − 4 = 2 x + 9 .

Solution: Both radicals are considered isolated on separate sides of the equation.

Answer: The solution is 13.

Example 8: Solve: x 2 + x − 14 3 = x + 50 3 .

Solution: Eliminate the radicals by cubing both sides.

Answer: The solutions are −8 and 8.

We will learn how to solve some of the more advanced radical equations in the next course, Intermediate Algebra.

Try this! Solve: 3 x + 1 = 2 x − 3 .

Video Solution

Key Takeaways

  • Solve equations involving square roots by first isolating the radical and then squaring both sides. Squaring a square root eliminates the radical, leaving us with an equation that can be solved using the techniques learned earlier in our study of algebra. However, squaring both sides of an equation introduces the possibility of extraneous solutions, so check your answers in the original equation.
  • Solve equations involving cube roots by first isolating the radical and then cubing both sides. This eliminates the radical and results in an equation that may be solved with techniques you have already mastered.

Topic Exercises

Part A: Solving Radical Equations

30. 9 ( x − 1 ) 3 = 3 ( x + 7 ) 3

71. 2 x 2 − 15 x + 25 = ( x + 5 ) ( x − 5 )

73. 2 ( x 2 + 3 x − 20 ) 3 = ( x + 3 ) 2 3

74. 3 x 2 + 3 x + 40 3 = ( x − 5 ) 2 3

83. ( 4 x + 15 ) 1 / 2 − 2 x = 0

87. 2 ( 5 x + 26 ) 1 / 2 = x + 10

89. The square root of 1 less than twice a number is equal to 2 less than the number. Find the number.

90. The square root of 4 less than twice a number is equal to 6 less than the number. Find the number.

91. The square root of twice a number is equal to one-half of that number. Find the number.

92. The square root of twice a number is equal to one-third of that number. Find the number.

93. The distance, d, measured in miles, a person can see an object is given by the formula d = 6 h 2

where h represents the person’s height above sea level, measured in feet. How high must a person be to see an object 5 miles away?

94. The current, I, measured in amperes, is given by the formula I = P R

where P is the power usage, measured in watts, and R is the resistance, measured in ohms. If a light bulb requires 1/2 ampere of current and uses 60 watts of power, then what is the resistance of the bulb?

The period, T, of a pendulum in seconds is given by the formula T = 2 π L 32

where L represents the length in feet. For each problem below, calculate the length of a pendulum, given the period. Give the exact value and the approximate value rounded off to the nearest tenth of a foot.

The time, t, in seconds an object is in free fall is given by the formula t = s 4

where s represents the distance in feet the object has fallen. For each problem below, calculate the distance an object falls, given the amount of time.

The x-intercepts for any graph have the form (x, 0), where x is a real number. Therefore, to find x-intercepts, set y = 0 and solve for x. Find the x-intercepts for each of the following.

107. Discuss reasons why we sometimes obtain extraneous solutions when solving radical equations. Are there ever any conditions where we do not need to check for extraneous solutions? Why?


Notice, we have $-2=sqrt<7-2b>-sqrt<2b+3>$ Squaring both the sides, we get $(-2)^2=(sqrt<7-2b>-sqrt<2b+3>)^2$ $4=7-2b+2b+3-2sqrt<7-2b>sqrt<2b+3>$ $sqrt<7-2b>sqrt<2b+3>=3$ Again, squaring both the sides, we get $(7-2b)(2b+3)=3^2$ $b^2-2b-3=0$$implies (b+1)(b-3)=0$

I hope you can solve further

You eliminate the radicals by squaring, repeating the operation as many times as necessary. The simplest, in general is to isolate a radical in one side.

However, beware the equation obtained after squaring is implied by the original equation, but not equivalent to it. Indeed the main rule is: $A=sqrt Biff A^2=Benspace color< ext>enspace Age 0$ Here, we first observe the domain of validity of the equation is $iggl[-dfrac32,dfrac72iggr]$.

We can rewrite the equation as egin sqrt<7-2b>+2&=sqrt<2b+3>Rightarrow 7-2b+4+4sqrt<7-2b>=2b+3iffsqrt<7-2b>=b-2&iff 7-2b=b^2-4b+4enspace color< ext>enspace bge 2 &iff b^2-2b-3=(b+1)(b-3)=0enspace color< ext>enspace bge 2 end Thus there is only one solution: $color$.

You can check the solution to the final equation, $b=-1$, is not a solution to the initial equation, as the right-hand side is equal to 2$ for this value of $b$.


Watch the video: Radical Equations (November 2021).