Exercise (PageIndex{1})

Simplify.

- (- sqrt { 121 })
- (sqrt { ( - 7 ) ^ { 2 } })
- (sqrt { ( x y ) ^ { 2 } })
- (sqrt { ( 6 x - 7 ) ^ { 2 } })
- (sqrt [ 3 ] { 125 })
- (sqrt [ 3 ] { - 27 })
- (sqrt [ 3 ] { ( x y ) ^ { 3 } })
- (sqrt [ 3 ] { ( 6 x + 1 ) ^ { 3 } })
- Given (f ( x ) = sqrt { x + 10 }), find (f(-1)) and (f(6)).
- Given (g(x) = sqrt [ 3 ] { x - 5 }), find (g(4)) and (g(13)).
- Determine the domain of the function defined by (g ( x ) = sqrt { 5 x + 2 }).
- Determine the domain of the function defined by (g ( x ) = sqrt [ 3 ] { 3 x - 1 }).

**Answer**1. (-11)

3. (|xy|)

5. (5)

7. (xy)

9. (f ( - 1 ) = 3 ; f ( 6 ) = 4)

11. (left[ - frac { 2 } { 5 } , infty ight))

Exercise (PageIndex{2})

Simplify.

- (sqrt [ 3 ] { 250 })
- (4 sqrt [ 3 ] { 120 })
- (- 3 sqrt [ 3 ] { 108 })
- (10 sqrt [ 5 ] { frac { 1 } { 32 } })
- (- 6 sqrt [ 4 ] { frac { 81 } { 16 } })
- (sqrt [ 6 ] { 128 })
- (sqrt [ 5 ] { - 192 })
- (- 3 sqrt { 420 })

**Answer**1. (5 sqrt [ 3 ] { 2 })

3. (- 9 sqrt [ 3 ] { 4 })

5. (-9)

7. (- 2 sqrt [ 5 ] { 6 })

Exercise (PageIndex{3})

Simplify.

- (sqrt { 20 x ^ { 4 } y ^ { 3 } })
- (- 4 sqrt { 54 x ^ { 6 } y ^ { 3 } })
- (sqrt { x ^ { 2 } - 14 x + 49 })
- (sqrt { ( x - 8 ) ^ { 4 } })

**Answer**1. (2 x ^ { 2 } | y | sqrt { 5 y })

3. (| x - 7 |)

Exercise (PageIndex{4})

Simplify. (Assume all variable expressions are nonzero.)

- (sqrt { 100 x ^ { 2 } y ^ { 4 } })
- (sqrt { 36 a ^ { 6 } b ^ { 2 } })
- (sqrt { frac { 8 a ^ { 2 } } { b ^ { 4 } } })
- (sqrt { frac { 72 x ^ { 4 } y } { z ^ { 6 } } })
- (10 x sqrt { 150 x ^ { 7 } y ^ { 4 } })
- (- 5 n ^ { 2 } sqrt { 25 m ^ { 10 } n ^ { 6 } })
- (sqrt [ 3 ] { 48 x ^ { 6 } y ^ { 3 } z ^ { 2 } })
- (sqrt [ 3 ] { 270 a ^ { 10 } b ^ { 8 } c ^ { 3 } })
- (sqrt [ 3 ] { frac { a ^ { 3 } b ^ { 5 } } { 64 c ^ { 6 } } })
- (sqrt [ 5 ] { frac { a ^ { 26 } } { 32 b ^ { 5 } c ^ { 10 } } })
- The period (T) in seconds of a pendulum is given by the formula (T = 2 pi sqrt { frac { L } { 32 } }) where (L) represents the length in feet of the pendulum. Calculate the period of a pendulum that is (2 frac{1}{2}) feet long. Give the exact answer and the approximate answer to the nearest hundredth of a second.
- The time in seconds an object is in free fall is given by the formula (t = frac { sqrt { s } } { 4 }) where (s) represents the distance in feet the object has fallen. How long does it take an object to fall (28) feet? Give the exact answer and the approximate answer to the nearest tenth of a second.
- Find the distance between ((−5, 6)) and ((−3,−4)).
- Find the distance between (left( frac { 2 } { 3 } , - frac { 1 } { 2 } ight)) and (left( 1 , - frac { 3 } { 4 } ight)).

**Answer**1. (10 x y ^ { 2 })

3. (frac { 2 a sqrt { 2 } } { b ^ { 2 } })

5. (50 x ^ { 4 } y ^ { 2 } sqrt { 6 x })

7. (2 x ^ { 2 } y sqrt [ 3 ] { 6 z ^ { 2 } })

9. (frac { a b sqrt [ 3 ] { b ^ { 2 } } } { 4 c ^ { 2 } })

11. (frac { pi sqrt { 5 } } { 4 }) seconds; (1.76) seconds

13. (2 sqrt { 26 }) units

Exercise (PageIndex{5})

Determine whether or not the three points form a right triangle. Use the Pythagorean theorem to justify your answer.

- (( - 4,5 ) , ( - 3 , - 1 ) , ext { and } ( 3,0 ))
- (( - 1 , - 1 ) , ( 1,3 ) , ext { and } ( - 6,1 ))

**Answer**1. Right triangle

Exercise (PageIndex{6})

Simplify. Assume all radicands containing variables are nonnegative.

- (7 sqrt { 2 } + 5 sqrt { 2 })
- (8 sqrt { 15 } - 2 sqrt { 15 })
- (14 sqrt { 3 } + 5 sqrt { 2 } - 5 sqrt { 3 } - 6 sqrt { 2 })
- (22 sqrt { a b } - 5 a sqrt { b } + 7 sqrt { a b } - 2 a sqrt { b })
- (7 sqrt { x } - ( 3 sqrt { x } + 2 sqrt { y } ))
- (( 8 y sqrt { x } - 7 x sqrt { y } ) - ( 5 x sqrt { y } - 12 y sqrt { x } ))
- (( 3 sqrt { 5 } + 2 sqrt { 6 } ) + ( 8 sqrt { 5 } - 3 sqrt { 6 } ))
- (( 4 sqrt [ 3 ] { 3 } - sqrt [ 3 ] { 12 } ) - ( 5 sqrt [ 3 ] { 3 } - 2 sqrt [ 3 ] { 12 } ))
- (( 2 - sqrt { 10 x } + 3 sqrt { y } ) - ( 1 + 2 sqrt { 10 x } - 6 sqrt { y } ))
- (left( 3 a sqrt [ 3 ] { a b ^ { 2 } } + 6 sqrt [ 3 ] { a ^ { 2 } b } ight) + left( 9 a sqrt [ 3 ] { a b ^ { 2 } } - 12 sqrt [ 3 ] { a ^ { 2 } b } ight))
- (sqrt { 45 } + sqrt { 12 } - sqrt { 20 } - sqrt { 75 })
- (sqrt { 24 } - sqrt { 32 } + sqrt { 54 } - 2 sqrt { 32 })
- (2 sqrt { 3 x ^ { 2 } } + sqrt { 45 x } - x sqrt { 27 } + sqrt { 20 x })
- (5 sqrt { 6 a ^ { 2 } b } + sqrt { 8 a ^ { 2 } b ^ { 2 } } - 2 sqrt { 24 a ^ { 2 } b } - a sqrt { 18 b ^ { 2 } })
- (5 y sqrt { 4 x ^ { 2 } y } - left( x sqrt { 16 y ^ { 3 } } - 2 sqrt { 9 x ^ { 2 } y ^ { 3 } } ight))
- (left( 2 b sqrt { 9 a ^ { 2 } c } - 3 a sqrt { 16 b ^ { 2 } c } ight) - left( sqrt { 64 a ^ { 2 } b ^ { 2 } c } - 9 b sqrt { a ^ { 2 } c } ight))
- (sqrt [ 3 ] { 216 x } - sqrt [ 3 ] { 125 x y } - sqrt [ 3 ] { 8 x })
- (sqrt [ 3 ] { 128 x ^ { 3 } } - 2 x sqrt [ 3 ] { 54 } + 3 sqrt [ 3 ] { 2 x ^ { 3 } })
- (sqrt [ 3 ] { 8 x ^ { 3 } y } - 2 x sqrt [ 3 ] { 8 y } + sqrt [ 3 ] { 27 x ^ { 3 } y } + x sqrt [ 3 ] { y })
- (sqrt [ 3 ] { 27 a ^ { 3 } b } - 3 sqrt [ 3 ] { 8 a b ^ { 3 } } + a sqrt [ 3 ] { 64 b } - b sqrt [ 3 ] { a })
- Calculate the perimeter of the triangle formed by the following set of vertices: ({ ( - 3 , - 2 ) , ( - 1,1 ) , ( 1 , - 2 ) }).
- Calculate the perimeter of the triangle formed by the following set of vertices: ({ ( 0 , - 4 ) , ( 2,0 ) , ( - 3,0 ) }).

**Answer**1. (12 sqrt { 2 })

3. (9 sqrt { 3 } - sqrt { 2 })

5. (4 sqrt { x } - 2 sqrt { y })

7. (11 sqrt { 5 } - sqrt { 6 })

9. (1 - 3 sqrt { 10 x } + 9 sqrt { y })

11. (sqrt { 5 } - 3 sqrt { 3 })

13. (- x sqrt { 3 } + 5 sqrt { 5 x })

15. (12 x y sqrt { y })

17. (4 sqrt [ 3 ] { x } - 5 sqrt [ 3 ] { x y })

19. (2 x sqrt [ 3 ] { y })

21. (4 + 2 sqrt { 13 }) units

Exercise (PageIndex{7})

Multiply.

- (sqrt { 6 } cdot sqrt { 15 })
- (( 4 sqrt { 2 } ) ^ { 2 })
- (sqrt { 2 } ( sqrt { 2 } - sqrt { 10 } ))
- (( sqrt { 5 } - sqrt { 6 } ) ^ { 2 })
- (( 5 - sqrt { 3 } ) ( 5 + sqrt { 3 } ))
- (( 2 sqrt { 6 } + sqrt { 3 } ) ( sqrt { 2 } - 5 sqrt { 3 } ))
- (( sqrt { a } - 5 sqrt { b } ) ^ { 2 })
- (3 sqrt { x y } ( sqrt { x } - 2 sqrt { y } ))
- (sqrt [ 3 ] { 3 a ^ { 2 } } cdot sqrt [ 3 ] { 18 a })
- (sqrt [ 3 ] { 49 a ^ { 2 } b } cdot sqrt [ 3 ] { 7 a ^ { 2 } b ^ { 2 } })

**Answer**1. (3 sqrt { 10 })

3. (2 - 2 sqrt { 5 })

5. (22)

7. (a - 10 sqrt { a b } + 25 b)

9. (3 a sqrt [ 3 ] { 2 })

Exercise (PageIndex{8})

Divide. Assume all variables represent nonzero numbers and rationalize the denominator where appropriate.

- (frac { sqrt { 72 } } { sqrt { 9 } })
- (frac { 10 sqrt { 48 } } { sqrt { 64 } })
- (frac { 5 } { sqrt { 5 } })
- (frac { sqrt { 15 } } { sqrt { 2 } })
- (frac { 3 } { 2 sqrt { 6 } })
- (frac { 2 + sqrt { 5 } } { sqrt { 10 } })
- (frac { 18 } { sqrt { 3 x } })
- (frac { 2 sqrt { 3 x } } { sqrt { 6 x y } })
- (frac { 1 } { sqrt [ 3 ] { 3 x ^ { 2 } } })
- (frac { 5 a b ^ { 2 } } { sqrt [ 3 ] { 5 a ^ { 2 } b } })
- (sqrt [ 3 ] { frac { 5 x z ^ { 2 } } { 49 x ^ { 2 } y ^ { 2 } z } })
- (frac { 1 } { sqrt [ 5 ] { 8 x ^ { 4 } y ^ { 2 } z } })
- (frac { 9 x ^ { 2 } y } { sqrt [ 5 ] { 81 x y ^ { 2 } z ^ { 3 } } })
- (sqrt [ 5 ] { frac { 27 a b ^ { 3 } } { 15 a ^ { 4 } b c ^ { 2 } } })
- (frac { 1 } { sqrt { 5 } - sqrt { 3 } })
- (frac { sqrt { 3 } } { sqrt { 2 } + 1 })
- (frac { - 3 sqrt { 6 } } { 2 - sqrt { 10 } })
- (frac { sqrt { x y } } { sqrt { x } - sqrt { y } })
- (frac { sqrt { 2 } - sqrt { 6 } } { sqrt { 2 } + sqrt { 6 } })
- (frac { sqrt { a } + sqrt { b } } { sqrt { a } - sqrt { b } })
- The base of a triangle measures (2 sqrt{6}) units and the height measures (3 sqrt{15}) units. Find the area of the triangle.
- If each side of a square measures (5+2 sqrt{10}) units, find the area of the square.

**Answer**1. (2 sqrt { 2 })

3. (sqrt { 5 })

5. (frac { sqrt { 6 } } { 4 })

7. (frac { 6 sqrt { 3 x } } { x })

9. (frac { sqrt [ 3 ] { 9 x } } { 3 x })

11. (frac { sqrt [ 3 ] { 35 x ^ { 2 } y z } } { 7 x y })

13. (frac { 3 x y sqrt [ 5 ] { 3 x ^ { 4 } y ^ { 3 } z ^ { 2 } } } { z })

15. (frac { sqrt { 5 } + sqrt { 3 } } { 2 })

17. (sqrt { 6 } + sqrt { 15 })

19. (- 2 + sqrt { 3 })

21. (9 sqrt { 10 }) square units

Exercise (PageIndex{9})

Express in radical form.

- (11 ^ { 1 / 2 })
- (2 ^ { 2 / 3 })
- (x ^ { 3 / 5 })
- (a ^ { - 4 / 5 })

**Answer**1. (sqrt { 11 })

3. (sqrt [ 5 ] { x ^ { 3 } })

Exercise (PageIndex{10})

Write as a radical and then simplify.

- (16 ^ { 1 / 2 })
- (72 ^ { 1 / 2 })
- (8 ^ { 2 / 3 })
- (32 ^ { 1 / 3 })
- (left( frac { 1 } { 9 } ight) ^ { 3 / 2 })
- (left( frac { 1 } { 216 } ight) ^ { - 1 / 3 })

**Answer**1. (4)

3. (4)

5. (frac{1}{27})

Exercise (PageIndex{11})

Perform the operations and simplify. Leave answers in exponential form.

- (6 ^ { 1 / 2 } cdot 6 ^ { 3 / 2 })
- (3 ^ { 1 / 3 } cdot 3 ^ { 1 / 2 })
- (frac { 6 ^ { 5 / 2 } } { 6 ^ { 3 / 2 } })
- (frac { 4 ^ { 3 / 4 } } { 4 ^ { 1 / 4 } })
- (left( 64 x ^ { 6 } y ^ { 2 } ight) ^ { 1 / 2 })
- (left( 27 x ^ { 12 } y ^ { 6 } ight) ^ { 1 / 3 })
- (left( frac { a ^ { 4 / 3 } } { a ^ { 1 / 2 } } ight) ^ { 2 / 5 })
- (left( frac { 16 x ^ { 4 / 3 } } { y ^ { 2 } } ight) ^ { 1 / 2 })
- (frac { 56 x ^ { 3 / 4 } y ^ { 3 / 2 } } { 14 x ^ { 1 / 2 } y ^ { 2 / 3 } })
- (frac { left( 4 a ^ { 4 } b ^ { 2 / 3 } c ^ { 4 / 3 } ight) ^ { 1 / 2 } } { 2 a ^ { 2 } b ^ { 1 / 6 } c ^ { 2 / 3 } })
- (left( 9 x ^ { - 4 / 3 } y ^ { 1 / 3 } ight) ^ { - 3 / 2 })
- (left( 16 x ^ { - 4 / 5 } y ^ { 1 / 2 } z ^ { - 2 / 3 } ight) ^ { - 3 / 4 })

**Answer**1. (36)

3. (6)

5. (8 x ^ { 3 } y)

7. (a ^ { 1 / 3 })

9. (4 x ^ { 1 / 4 } y ^ { 5 / 6 })

11. (frac { x ^ { 2 } } { 27 y ^ { 1 / 2 } })

Exercise (PageIndex{12})

Perform the operations with mixed indices.

- (sqrt { y } cdot sqrt [ 5 ] { y ^ { 2 } })
- (sqrt [ 3 ] { y } cdot sqrt [ 5 ] { y ^ { 3 } })
- (frac { sqrt [ 3 ] { y ^ { 2 } } } { sqrt [ 3 ] { y } })
- (sqrt { sqrt [ 3 ] { y ^ { 2 } } })

**Answer**1. (sqrt [ 10 ] { y ^ { 9 } })

3. (sqrt [ 15 ] { y ^ { 7 } })

Exercise (PageIndex{13})

Solve.

- (2 sqrt { x } + 3 = 13)
- (sqrt { 3 x - 2 } = 4)
- (sqrt { x - 5 } + 4 = 8)
- (5 sqrt { x + 3 } + 7 = 2)
- (sqrt { 4 x - 3 } = sqrt { 2 x + 15 })
- (sqrt { 8 x - 15 } = x)
- (x - 1 = sqrt { 13 - x })
- (sqrt { 4 x - 3 } = 2 x - 3)
- (sqrt { x + 5 } = 5 - sqrt { x })
- (sqrt { x + 3 } = 3 sqrt { x } - 1)
- (sqrt { 2 ( x + 1 ) } - sqrt { x + 2 } = 1)
- (sqrt { 6 - x } + sqrt { x - 2 } = 2)
- (sqrt { 3 x - 2 } + sqrt { x - 1 } = 1)
- (sqrt { 9 - x } = sqrt { x + 16 } - 1)
- (sqrt [ 3 ] { 4 x - 3 } = 2)
- (sqrt [ 3 ] { x - 8 } = - 1)
- (sqrt [ 3 ] { x ( 3 x + 10 ) } = 2)
- (sqrt [ 3 ] { 2 x ^ { 2 } - x } + 4 = 5)
- (sqrt [ 3 ] { 3 ( x + 4 ) ( x + 1 ) } = sqrt [ 3 ] { 5 x + 37 })
- (sqrt [ 3 ] { 3 x ^ { 2 } - 9 x + 24 } = sqrt [ 3 ] { ( x + 2 ) ^ { 2 } })
- (y ^ { 1 / 2 } - 3 = 0)
- (y ^ { 1 / 3 } + 3 = 0)
- (( x - 5 ) ^ { 1 / 2 } - 2 = 0)
- (( 2 x - 1 ) ^ { 1 / 3 } - 5 = 0)
- (( x - 1 ) ^ { 1 / 2 } = x ^ { 1 / 2 } - 1)
- (( x - 2 ) ^ { 1 / 2 } - ( x - 6 ) ^ { 1 / 2 } = 2)
- (( x + 4 ) ^ { 1 / 2 } - ( 3 x ) ^ { 1 / 2 } = - 2)
- (( 5 x + 6 ) ^ { 1 / 2 } = 3 - ( x + 3 ) ^ { 1 / 2 })
- Solve for (g : t = sqrt { frac { 2 s } { g } }).
- Solve for (x:y = sqrt [ 3 ] { x + 4 } - 2),
- The period in seconds of a pendulum is given by the formula (T = 2 pi sqrt { frac { L } { 32 } }) where (L) represents the length in feet of the pendulum. Find the length of a pendulum that has a period of (1 frac{1}{2}) seconds. Find the exact answer and the approximate answer rounded off to the nearest tenth of a foot.
- The outer radius of a spherical shell is given by the formula (r = sqrt [ 3 ] { frac { 3 V } { 4 pi } } + 2) where (V) represents the inner volume in cubic centimeters. If the outer radius measures (8) centimeters, find the inner volume of the sphere.
- The speed of a vehicle before the brakes are applied can be estimated by the length of the skid marks left on the road. On dry pavement, the speed (v) in miles per hour can be estimated by the formula (v = 2 sqrt { 6 d }), where (d) represents the length of the skid marks in feet. Estimate the length of a skid mark if the vehicle is traveling (30) miles per hour before the brakes are applied.
- Find the real root of the function defined by (f ( x ) = sqrt [ 3 ] { x - 3 } + 2).

**Answer**1. (25)

3. (21)

5. (9)

7. (4)

9. (4)

11. (7)

13. (1)

15. (frac{11}{4})

17. (−4, frac{2}{3})

19. (−5, frac{5}{3})

21. (9)

23. (9)

25. (1)

27. (12)

29. (g = frac { 2 s } { t ^ { 2 } })

31. (frac { 18 } { pi ^ { 2 } }) feet; (1.8) feet

33. (37.5) feet

Exercise (PageIndex{14})

Write the complex number in standard form (a+bi).

- (5 - sqrt { - 16 })
- (- sqrt { - 25 } - 6)
- (frac { 3 + sqrt { - 8 } } { 10 })
- (frac { sqrt { - 12 } - 4 } { 6 })

**Answer**1. (5 - 4 i)

3. (frac { 3 } { 10 } + frac { sqrt { 2 } } { 5 } i)

Exercise (PageIndex{15})

Perform the operations.

- (( 6 - 12 i ) + ( 4 + 7 i ))
- (( - 3 + 2 i ) - ( 6 - 4 i ))
- (left( frac { 1 } { 2 } - i ight) - left( frac { 3 } { 4 } - frac { 3 } { 2 } i ight))
- (left( frac { 5 } { 8 } - frac { 1 } { 5 } i ight) + left( frac { 3 } { 2 } - frac { 2 } { 3 } i ight))
- (( 5 - 2 i ) - ( 6 - 7 i ) + ( 4 - 4 i ))
- (( 10 - 3 i ) + ( 20 + 5 i ) - ( 30 - 15 i ))
- (4 i ( 2 - 3 i ))
- (( 2 + 3 i ) ( 5 - 2 i ))
- (( 4 + i ) ^ { 2 })
- (( 8 - 3 i ) ^ { 2 })
- (( 3 + 2 i ) ( 3 - 2 i ))
- (( - 1 + 5 i ) ( - 1 - 5 i ))
- (frac { 2 + 9 i } { 2 i })
- (frac { i } { 1 - 2 i })
- (frac { 4 + 5 i } { 2 - i })
- (frac { 3 - 2 i } { 3 + 2 i })
- (10 - 5 ( 2 - 3 i ) ^ { 2 })
- (( 2 - 3 i ) ^ { 2 } - ( 2 - 3 i ) + 4)
- (left( frac { 1 } { 1 - i } ight) ^ { 2 })
- (left( frac { 1 + 2 i } { 3 i } ight) ^ { 2 })
- (sqrt { - 8 } ( sqrt { 3 } - sqrt { - 4 } ))
- (( 1 - sqrt { - 18 } ) ( 3 - sqrt { - 2 } ))
- (( sqrt { - 5 } - sqrt { - 10 } ) ^ { 2 })
- (( 1 - sqrt { - 2 } ) ^ { 2 } - ( 1 + sqrt { - 2 } ) ^ { 2 })
- Show that both (-5i) and (5i) satisfy (x^{2}+25=0).
- Show that both (1-2i) and (1+2i) satisfy (x^{2}-2x+5=0).

**Answer**1. (10 - 5 i)

3. (- frac { 1 } { 4 } + frac { 1 } { 2 } i)

5. (3+i)

7. (12+8i)

9. (15+8i)

11. (13)

13. (frac{9}{2}-i)

15. (frac { 3 } { 5 } + frac { 14 } { 5 } i)

17. (35+60i)

19. (frac{1}{2}i)

21. (4 sqrt { 2 } + 2 i sqrt { 6 })

23. (- 15 + 10 sqrt { 2 })

25. Answer may vary

## Sample Exam

Exercise (PageIndex{16})

Simplify. (Assume all variables are positive.)

- (5 x sqrt { 121 x ^ { 2 } y ^ { 4 } })
- (2 x y ^ { 2 } sqrt [ 3 ] { - 64 x ^ { 6 } y ^ { 9 } })
- Calculate the distance between ((-5,-3)) and ((-2,6)).
- The time in seconds an object is in free fall is given by the formula (t = frac { sqrt { s } } { 4 }) where (s) represents the distance in feet that the object has fallen. If a stone is dropped into a (36)-foot pit, how long will it take to hit the bottom of the pit?

**Answer**1. (55 x ^ { 2 } y ^ { 2 })

3. (3sqrt{10}) units

Exercise (PageIndex{17})

Perform the operations and simplify. (Assume all variables are positive and rationalize the denominator where appropriate.)

- (sqrt { 150 x y ^ { 2 } } - 2 sqrt { 18 x ^ { 3 } } + y sqrt { 24 x } + x sqrt { 128 x })
- (3 sqrt [ 3 ] { 16 x ^ { 3 } y ^ { 2 } } - left( 2 x sqrt [ 3 ] { 250 y ^ { 2 } } - sqrt [ 3 ] { 54 x ^ { 3 } y ^ { 2 } } ight))
- (2 sqrt { 2 } ( sqrt { 2 } - 3 sqrt { 6 } ))
- (( sqrt { 10 } - sqrt { 5 } ) ^ { 2 })
- (frac { sqrt { 6 } } { sqrt { 2 } + sqrt { 3 } })
- (frac { 2 x } { sqrt { 2 x y } })
- (frac { 1 } { sqrt [ 5 ] { 8 x y ^ { 2 } z ^ { 4 } } })
- Simplify: (81 ^ { 3 / 4 }).
- Express in radical form: (x ^ { - 3 / 5 }).

**Answer**1. (7 y sqrt { 6 x } + 2 x sqrt { 2 x })

3. (4 - 12 sqrt { 3 })

5. (- 2 sqrt { 3 } + 3 sqrt { 2 })

7. (frac { sqrt [ 5 ] { 4 x ^ { 4 } y ^ { 3 } z } } { 2 x y z })

9. (frac { 1 } { sqrt [ 5 ] { x ^ { 3 } } })

Exercise (PageIndex{18})

Solve.

- (sqrt { x } - 5 = 1)
- (sqrt [ 3 ] { 5 x - 2 } + 6 = 4)
- (5 sqrt { 2 x + 5 } - 2 x = 11)
- (sqrt { 4 - 3 x } + 2 = x)
- (sqrt { 2 x + 5 } - sqrt { x + 3 } = 2)
- The time in seconds an object is in free fall is given by the formula (t = frac { sqrt { s } } { 4 }) where (s) represents the distance in feet that the object has fallen. If a stone is dropped into a pit and it takes (4) seconds to reach the bottom, how deep is the pit?
- The width in inches of a container is given by the formula (w = frac { sqrt [ 3 ] { 4 V } } { 2 } + 1) where (V) represents the inside volume in cubic inches of the container. What is the inside volume of the container if the width is (6) inches?

**Answer**2. (-frac{6}{5})

4. (varnothing)

6. (256) feet

Exercise (PageIndex{19})

Perform the operations and write the answer in standard form.

- (sqrt { - 3 } ( sqrt { 6 } - sqrt { - 3 } ))
- (frac { 4 + 3 i } { 2 - i })
- (6 - 3 ( 2 - 3 i ) ^ { 2 })

**Answer**1. (3 + 3 i sqrt { 2 })

3. (21 + 36 i)

## RADICAL EQUATIONS WITH EXTRANEOUS SOLUTIONS WORKSHEET

**Radical equations with extraneous solutions worksheet :**

Here we are going to see some practice questions on radical equations with extraneous solutions.

What is extraneous solution ?

An extraneous solution is a solution derived from an equation that is not a solution of the original equation.

Therefore, you must check all solutions in the original equation when you solve radical equations.

## Radicals Worksheets

Let students get instant access to our free printable assortment of radicals worksheets, so they quickly work around their difficulties understanding the parts of a radical, simplifying a radical expression, and performing the four basic arithmetic operations with radicals. These exclusive exercises are a welcome opportunity for youngsters to practice rationalizing the denominator of a fraction and finding square roots and cube roots of numerals using prime factorization.

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Help students get accustomed to finding the square root and cube root of numbers with this free radical worksheet. Certain radicands presented here are neither perfect cubes nor perfect squares. Get oodles of practice simplifying such radicals too.

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Be conversant with the basic arithmetic operations: addition, subtraction, multiplication, and division involving radicals with this worksheet pdf. Simplify the radicals wherever necessary.

## One-step Equations Worksheets

Get introduced to the world of one-step equations with these free printable worksheets. Heighten your equation-solving skills by practicing one-step equations involving addition, subtraction, multiplication and division. An ensemble of worksheets with exercises like solving one-step equations involving integers, fractions, and decimals are incorporated here.

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Revise and review the concept with this pdf solving one-step equations worksheet! Get your students to plug the values in the equation and verify the solutions too.

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## Algebra 2

You will learn about Numbers, Polynomials, Inequalities, Sequences and Sums, many types of Functions, and how to **solve** them.

You will also gain a deeper insight into Mathematics, get to practice using your new skills with lots of examples and questions, and generally improve your mind.

With your new skills you will be able to put together mathematical models so you can find good quality solutions to many tricky real world situations.

Near the end of most pages is a "Your Turn" section . do these! You need to balance your reading with **doing**. Answering questions helps you sort things out in your mind. And don't guess the answer: use pen and paper and try your best before seeing the solution.

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**Math workbook 1** is a content-rich downloadable **zip** file with **100 Math printable exercises** and 100 pages of answer sheets attached to each exercise. This product is suitable for **Preschool, kindergarten** and **Grade 1**. The product is available for instant download after purchase. Not convinced? Take a video tour by clicking on the e-book icon to your left.

## 5.E: Radical Functions and Equations (Exercises) - Mathematics

### Proving triangles congruent

Two triangles are congruent if all six parts have the same measures. The three angles and the three sides must match. There is, however, a shorter way to prove that two triangles are congruent! In some cases, we are allowed to say that two triangles are congruent if a certain 3 parts match because the other 3 MUST be the same because of it. There are five of these certain cases and they are called *postulates*, which basically just means a rule.

**1. Side-Side-Side (SSS)**

If we know that the three sides of a triangle are congruent to the three sides of another triangle, then the angles MUST be the same (or it wouldn’t form a triangle).

( herefore Delta ABC cong Delta DEF)

The symbol ( herefore ) means “therefore.” If we are able to show that the three corresponding sides are congruent, then we have enough information to prove that the two triangles are congruent because of the __SSS Postulate!__

**2. Side-Angle-Side (SAS)**

If we can show that two sides and the angle IN BETWEEN them are congruent, then the whole triangle must be congruent as well. It looks like this:

(angle B cong angle E) (*angle*)

( herefore Delta ABC cong Delta DEF)

The angle HAS to be in between the two sides for the __SAS Postulate__ to be used.

**3. Angle-Side-Angle (ASA)**

If we can show that two angles and the side IN BETWEEN them are congruent, then the whole triangle must be congruent as well.

(angle B cong angle E) (*angle*)

(angle C cong angle F) (*angle*)

( herefore Delta ABC cong Delta DEF)

The side HAS to be in between the two angles for the __ASA Postulate__ to be used.

**4. Angle-Angle-Side (AAS)**

If the side is not in between the angles, it’s actually still okay, but you must use the __AAS Postulate__.

(angle B cong angle E) (*angle*)

(angle C cong angle F) (*angle*)

( herefore Delta ABC cong Delta DEF)

**5. Hypotenuse-Leg (HL)**

This one is a little bit different. It can only be used in a right triangle. So, if the two triangles are both right triangles and one of their corresponding legs are congruent as well as their hypotenuse, then they are congruent by the __HL Postulate__.

(Delta QRS& Delta XYZ) are right triangles

(overline *hypotenuse*)

( herefore Delta QRS cong Delta XYZ)

If you don’t have one of these above postulates, then you don’t have enough information to assume that the triangles are congruent.

You CANNOT use Angle-Angle-Angle because two triangles can have the same angles but have completely different sides, like the following example:

**Not Congruent!**

You also CANNOT use Side-Side-Angle because it doesn’t guarantee that the triangles will be completely congruent.

**Not Congruent!**

Below you can **download** some **free** math worksheets and practice.

## 5.E: Radical Functions and Equations (Exercises) - Mathematics

· Given a function, described by an equation, find function values (outputs) for specified inputs.

Throughout this course, you have been working with algebraic equations. Many of these equations are functions. For example, *y* = 4*x* +1 is an equation that represents a function. When you input values for *x*, you can determine a single output for *y*. In this case, if you substitute *x* = 10 into the equation you will find that *y* must be 41 there is no other value of *y* that would make the equation true.

Rather than using the variable *y,* the equations of functions can be written using **function notation**. Function notation is very useful when you are working with more than one function at a time, and substituting more than one variable in for *x.*

Some people think of functions as “mathematical machines.” Imagine you have a machine that changes a number according to a specific rule, such as “multiply by 3 and add 2” or “divide by 5, add 25, and multiply by −1.” If you put a number into the machine, a new number will pop out the other end, having been changed according to the rule. The number that goes in is called the input, and the number that is produced is called the output.

You can also call the machine “*f”* for function. If you put *x* into the box, *f*(*x*)*,*comes out. Mathematically speaking, *x* is the input, or the “independent variable,” and *f*(*x*) is the output, or the “dependent variable,” since it depends on the value of *x*.

*f* (*x*)*=* 4*x* + 1 is written in function notation and is read “*f* of *x* equals 4*x* plus 1.” It represents the following situation: A function named *f* acts upon an input, *x,* and produces *f*(*x*) which is equal to 4*x* + 1. This is the same as the equation as *y* = 4*x* + 1.

Function notation gives you more flexibility because you don’t have to use *y* for every equation. Instead, you could use *f*(*x*) or *g*(*x*) or *c*(*x*). This can be a helpful way to distinguish equations of functions when you are dealing with more than one at a time.

You could write the formula for perimeter, *P* = 4*s*, as the function *p*(*x*) = 4*x*, and the formula for area, *A* = *x* 2 , as *a*(*x*) = *x* 2 . This would make it easy to graph both functions on the same graph without confusion about the variables.

Which two equations represent the same function?

A) *y* = 2*x* – 7 and *f*(*x*) = 7 – 2*x*

B) 3*x* = *y* – 2 and *f*(*x*) = 3*x* – 2

C) *f*(*x*) = 3*x* 2 + 5 and *y* = 3*x* 2 + 5

A) *y* = 2*x* – 7 and *f*(*x*) = 7 – 2*x*

Incorrect. These equations look similar but are not the same. The first has a slope of 2 and a *y*-intercept of −7. The second function has a slope of −2 and a *y*-intercept of 7. It slopes in the opposite direction. They do not produce the same graph, so they are not the same function. The correct answer is *f*(*x*) = 3*x* 2 + 5 and *y* = 3*x* 2 + 5.

B) 3*x* = *y* – 2 and *f*(*x*) = 3*x* – 2

Incorrect. These equations represent two different functions. If you rewrite the first equation in terms of *y,* you’ll find the equation of the function is *y* = 3*x +* 2. The correct answer is *f*(*x*) = 3*x* 2 + 5 and *y* = 3*x* 2 + 5.

C) *f*(*x*) = 3*x* 2 + 5 and *y* = 3*x* 2 + 5

Correct. The expressions that follow *f*(*x*) = and *y* = are the same, so these are two different ways to write the same function: *f*(*x*) = 3*x* 2 + 5 and *y* = 3*x* 2 + 5.

Incorrect. Look at the expressions that follow *f*(*x*) = and *y* =. If the expressions are the same, then the equations represent the same exact function. The correct answer is *f*(*x*) = 3*x* 2 + 5 and *y* = 3*x* 2 + 5.

Equations written using function notation can also be evaluated. With function notation, you might see a problem like this.

*Given f* (*x*) *= 4x + 1, find f*(*2*)*.*

You read this problem like this: “given *f* of *x* equals 4*x* plus one, find *f* of 2.” While the notation and wording is different, the process of evaluating a function is the same as evaluating an equation: in both cases, you substitute 2 for *x*, multiply it by 4 and add 1, simplifying to get 9. In both a function and an equation, an input of 2 results in an output of 9.

*f*(*x*) = 4*x* + 1

You can simply apply what you already know about evaluating expressions to evaluate a function. It’s important to note that the parentheses that are part of function notation do not mean multiply. The notation *f*(*x*) does not mean *f* multiplied by *x*. Instead the notation means “*f* of *x*” or “the function of *x*” To evaluate the function, take the value given for *x* , and substitute that value in for *x* in the expression. Let’s look at a couple of examples.

**Given f(x) = 3x – 4, find f(5).**

## 5.E: Radical Functions and Equations (Exercises) - Mathematics

Recall that a real number can be interpreted as the measure of the angle constructed as follows: wrap a piece of string of length units around the unit circle (counterclockwise if , clockwise if ) with initial point P (1,0) and terminal point Q ( x , y ). This gives rise to the central angle with vertex O (0,0) and sides through the points P and Q . All six trigonometric functions of are defined in terms of the coordinates of the point Q ( x , y ), as follows:

Since Q ( x , y ) is a point on the unit circle, we know that . This fact and the definitions of the trigonometric functions give rise to the following fundamental identities:

This modern notation for trigonometric functions is due to L. Euler (1748).

More generally, if Q ( x , y ) is the point where the circle of radius R is intersected by the angle , then it follows (from similar triangles) that

If an angle corresponds to a point Q ( x , y ) on the unit circle, it is not hard to see that the angle corresponds to the same point Q ( x , y ), and hence that

Moreover, is the smallest positive angle for which Equations 1 are true for any angle . In general, we have for all angles :

We call the number the period of the trigonometric functions and , and refer to these functions as being periodic . Both and are periodic functions as well, with period , while and are periodic with period .

EXAMPLE 1 Find the period of the function .

Solution: The function runs through a full cycle when the angle 3 x runs from 0 to , or equivalently when x goes from 0 to . The period of f ( x ) is then .

EXERCISE 1 Find the period of the function .

Evaluation of Trigonometric functions

Consider the triangle with sides of length and hypotenuse c >0 as in Figure 1 below:

For the angle pictured in the figure, we see that

There are a few angles for which all trigonometric functions may be found using the triangles shown in the following Figure 2.

This list may be extended with the use of reference angles (see Example 2 below).

EXAMPLE 1: Find the values of all trigonometric functions of the angle .

Solution: From Figure 2, we see that the angle of corresponds to the point on the unit circle, and so

EXAMPLE 2: Find the values of all trigonometric functions of the angle .

Solution: Observe that an angle of is equivalent to 8 whole revolutions (a total of ) plus , Hence the angles and intersect the unit circle at the same point Q ( x , y ), and so their trigonometric functions are the same. Furthermore, the angle of makes an angle of with respect to the x-axis (in the second quadrant). From this we can see that and hence that

We call the auxiliary angle of the reference angle of .

**EXAMPLE 3** Find all trigonometric functions of an angle in the third quadrant for which .

Solution: We first construct a point R ( x , y ) on the terminal side of the angle , in the third quadrant. If R ( x , y ) is such a point, then and we see that we may take x =-5 and R =6. Since we find that (the negative signs on x and y are taken so that R ( x , y ) is a point on the third quadrant, see Figure 3).

Here are some Exercises on the evaluation of trigonometric functions.

EXERCISE 2 (a) Evaluate (give the exact answer).

(b) If and , find (give the **exact** answer).

EXERCISE 3 From a 200-foot observation tower on the beach, a man sights a whale in difficulty. The angle of depression of the whale is . How far is the whale from the shoreline?

## 5.E: Radical Functions and Equations (Exercises) - Mathematics

Click HERE to see a detailed solution to problem 1.

Click HERE to see a detailed solution to problem 2.

Click HERE to see a detailed solution to problem 3.

Click HERE to see a detailed solution to problem 4.

Click HERE to see a detailed solution to problem 5.

Click HERE to see a detailed solution to problem 6.

Click HERE to see a detailed solution to problem 7.

Click HERE to see a detailed solution to problem 8.

Click HERE to see a detailed solution to problem 9.

Click HERE to see a detailed solution to problem 10.

Click HERE to see a detailed solution to problem 11.

Click HERE to see a detailed solution to problem 12.

This problem requires an unusual replacement, trigonometry identities, and trigonometry limits.

### The next problem requires an understanding of one-sided limits.

ii.) Determine the following limits.

Click HERE to see a detailed solution to problem 14.

Determine the values of constants a and b so that exists and is equal to f(2).

### Click HERE to return to the original list of various types of calculus problems.

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