Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a...

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Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 = 81 • 3 81 is a perfect square and a factor of 243. 81 • 3 Use the Multiplication Property of Square Roots. 9 3 Simplify 81. Quick Check 11-1

Transcript of Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a...

Page 1: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Simplify 243.

Simplifying RadicalsSimplifying RadicalsALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

243 = 81 • 3 81 is a perfect square and a factor of 243.

= 81 • 3 Use the Multiplication Property of Square Roots.

= 9 3 Simplify 81.

Quick Check

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Page 2: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Simplify 28x7.

Simplifying RadicalsSimplifying RadicalsALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

28x7 = 4x6 • 7x 4x6 is a perfect square and a factor of 28x7.

= 4x6 • 7x Use the Multiplication Property of Square Roots.

= 2x3 7x Simplify 4x6.

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Page 3: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Simplifying RadicalsSimplifying Radicals

Simplify each radical expression.

ALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

a. 12 • 32 12 • 32 = 12 • 32 Use the Multiplication Property of

Square Roots.

= 384 Simplify under the radical.

= 64 • 6 64 is a perfect square and a factor of 384.

= 64 • 6 Use the Multiplication Property of

Square Roots.

= 8 6 Simplify 64.

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Page 4: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Simplifying RadicalsSimplifying Radicals

(continued)

ALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

b. 7 5x • 3 8x

= 42x 10 Simplify.

= 21 • 2x 10 Simplify 4x2.

= 21 4x2 • 10 Use the Multiplication Property of

Square Roots.

= 21 4x2 • 10 4x2 is a perfect square and a

factor of 40x2.

7 5x • 3 8x = 21 40x2 Multiply the whole numbers and

use the Multiplication Property of

Square Roots.

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Page 5: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Simplifying RadicalsSimplifying Radicals

Suppose you are looking out a fourth floor window 52 ft above

the ground. Use the formula d = 1.5h to estimate the distance you

can see to the horizon. Round your answer to the nearest mile.

ALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

d = 1.5h

To the nearest mile, the distance you can see is 9 miles.

= 78 Multiply.

= 1.5 • 52 Substitute 52 for h.

8.83176 Use a calculator.

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Page 6: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Simplifying RadicalsSimplifying Radicals

Simplify each radical expression.

ALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

= Simplify 64. 13

8

a. 1364

b. 49x4

7

x2 = Simplify 49 and x4.

= Use the Division Property of Square Roots.1364

13

64

= Use the Division Property of Square Roots.49x4

49

x4

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Page 7: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Simplifying RadicalsSimplifying Radicals

= 12 Divide.120 10

= 4 • 3 4 is a perfect square and a factor of 12.

a. 120 10

Simplify each radical expression.

= 4 • 3 Use the Multiplication Property of Square Roots.

= 2 3 Simplify 4.

ALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

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Page 8: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

b. 75x5

48x

Simplifying RadicalsSimplifying Radicals

= Divide the numerator and denominator by 3x.75x5

48x25x4

16

= Use the Division Property of Square Roots.25x4

16

(continued)

= Use the Multiplication Property ofSquare Roots.

25 • x4

16

= Simplify 25, x4, and 16.5x2

4

ALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

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Page 9: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

3

7

3

7

7

7 7

7= • Multiply by to make the denominator a

perfect square.

Simplifying RadicalsSimplifying Radicals

Simplify each radical expression.

a. 3 7

= Simplify 49.3 7 7

= Use the Multiplication Property of Square Roots.3 7

49

ALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

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Page 10: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

= Simplify 36x4. 33x

6x2

Simplifying RadicalsSimplifying Radicals

(continued)

b. 11

12x3

Simplify the radical expression.

= • Multiply by to make the denominator a

perfect square.

3x

3x

3x

3x

11

12x3

11

12x3

= Use the Multiplication Property of Square Roots. 33x

36x4

ALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

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Page 11: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Simplifying RadicalsSimplifying RadicalsALGEBRA 1 LESSON 11-1ALGEBRA 1 LESSON 11-1

12

36

Simplify each radical expression.

1. 16 • 8 2. 4 144 3.

4. 5. 2

a5

3x

15x3

8 2 48 3

3

2 a a3

5 5x

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Page 12: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Simplify 4 3 + 3.

Operations with Radical ExpressionsOperations with Radical ExpressionsALGEBRA 1 LESSON 11-2ALGEBRA 1 LESSON 11-2

= (4 + 1) 3 Use the Distributive Property to combine like radicals.

= 5 3 Simplify.

4 3 + 3 = 4 3 + 1 3 Both terms contain 3.

Quick Check

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Page 13: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

8 5 – 45 = 8 5 + 9 • 5 9 is a perfect square and a factor of 45.

Simplify 8 5 – 45.

Operations with Radical ExpressionsOperations with Radical ExpressionsALGEBRA 1 LESSON 11-2ALGEBRA 1 LESSON 11-2

= 8 5 – 9 • 5 Use the Multiplication Property of Square Roots.

= 8 5 – 3 5 Simplify 9.

= (8 – 3) 5 Use the Distributive Property tocombine like terms.

= 5 5 Simplify.

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Page 14: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Operations with Radical ExpressionsOperations with Radical Expressions

Simplify 5( 8 + 9).

ALGEBRA 1 LESSON 11-2ALGEBRA 1 LESSON 11-2

5( 8 + 9) = 40 + 9 5 Use the Distributive Property.

= 4 • 10 + 9 5 Use the Multiplication Property of Square Roots.

= 2 10 + 9 5 Simplify.

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Page 15: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Simplify ( 6 – 3 21)( 6 + 21).

Operations with Radical ExpressionsOperations with Radical ExpressionsALGEBRA 1 LESSON 11-2ALGEBRA 1 LESSON 11-2

( 6 – 3 21)( 6 + 21)

= 36 + 126 – 3 126 – 3 441 Use

FOIL.= 6 – 2 126 – 3(21) Combine like radicals and

simplify 36 and 441.

= 6 – 2 9 • 14 – 63 9 is a perfect square factor of 126.

= 6 – 2 9 • 14 – 63 Use the Multiplication Property of Square Roots.

= 6 – 6 14 – 63 Simplify 9.

= –57 – 6 14 Simplify.

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Page 16: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Operations with Radical ExpressionsOperations with Radical ExpressionsALGEBRA 1 LESSON 11-2ALGEBRA 1 LESSON 11-2

= 2( 7 + 3) Divide 8 and 4 by the common factor 4.

= 2 7 + 2 3 Simplify the expression.

= Multiply in the denominator. 8( 7 + 3)

7 – 3

= Simplify the denominator. 8( 7 + 3)

4

Simplify . 8

7 – 3

= • Multiply the numerator and denominator by the conjugate of the denominator.

8

7 – 3

7 + 3

7 + 3

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Page 17: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Solve each equation. Check your answers.

Solving Radical EquationsSolving Radical EquationsALGEBRA 1 LESSON 11-3ALGEBRA 1 LESSON 11-3

a. x – 5 = 4

x = 9 Isolate the radical on the left side of the equation.

( x)2 = 92 Square each side.

x = 81

x – 5 = 4

4 = 4

Check: x – 5 = 4

81 – 5 4 Substitute 81 for x.

9 – 5 4

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Page 18: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

b. x – 5 = 4

Solving Radical EquationsSolving Radical EquationsALGEBRA 1 LESSON 11-3ALGEBRA 1 LESSON 11-3

x – 5 = 16 Solve for x.

x = 21

( x – 5)2 = 42 Square each side.

(continued)

Check: x – 5 = 4 21– 5 = 4 Substitute 21 for x. 16 = 4

4 = 4

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Page 19: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

On a roller coaster ride, your speed in a loop depends on the

height of the hill you have just come down and the radius of the loop in

feet. The equation v = 8 h – 2r gives the velocity v in feet per second

of a car at the top of the loop.

Solving Radical EquationsSolving Radical EquationsALGEBRA 1 LESSON 11-3ALGEBRA 1 LESSON 11-3

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Page 20: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Solving Radical EquationsSolving Radical EquationsALGEBRA 1 LESSON 11-3ALGEBRA 1 LESSON 11-3

The loop on a roller coaster ride has a radius of 18 ft. Your car has a velocity of 120 ft/s at the top of the loop. How high is the hill of the loop you have just come down before going into the loop?

Solve v = 8 h – 2r for h when v = 120 and r = 18.120 = 8 h – 2(18) Substitute 120 for v and 18 for r.

= Divide each side by 8 to isolate the radical.

15 = h – 36 Simplify.

8 h – 2(18) 8

120 8

(15)2 = ( h – 36)2 Square both sides.225 = h – 36261 = h

The hill is 261 ft high.

(continued)

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Page 21: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Solve 3x – 4 = 2x + 3.

Solving Radical EquationsSolving Radical Equations

( 3x – 4)2 = ( 2x + 3)2 Square both sides.

3x – 4 = 2x + 3 Simplify.

3x = 2x + 7 Add 4 to each side.

x = 7 Subtract 2x from each side.

The solution is 7.

Check: 3x – 4 = 2x + 3

3(7) – 4 2(7) + 3 Substitute 7 for x.

17 = 17

ALGEBRA 1 LESSON 11-3ALGEBRA 1 LESSON 11-3

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Page 22: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Solving Radical EquationsSolving Radical EquationsALGEBRA 1 LESSON 11-3ALGEBRA 1 LESSON 11-3

(x)2 = ( x + 12)2 Square both sides.

x2 = x + 12

x2 – x – 12 = 0 Simplify.

The solution to the original equation is 4. The value –3 is an extraneous solution.

Solve x = x + 12.

(x – 4)(x + 3) = 0 Solve the quadratic equation by factoring.

(x – 4) = 0 or (x + 3) = 0 Use the Zero–Product Property. x = 4  or   x = –3 Solve for x.

Check: x = x + 12

4 4 + 12 –3 –3 + 12

4 = 4 –3 = 3 /

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Page 23: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Solving Radical EquationsSolving Radical EquationsALGEBRA 1 LESSON 11-3ALGEBRA 1 LESSON 11-3

Solve 3x + 8 = 2.

3x = –6

( 3x)2 = (–6)2 Square both sides.

3x = 36

x = 12

3x + 8 = 2 has no solution.

Check: 3x + 8 = 2

3(12) + 8 2   Substitute 12 for x.

36 + 8 2

6 + 8 = 2   x = 12 does not solve the original equation./

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Page 24: Simplify 243. Simplifying Radicals ALGEBRA 1 LESSON 11-1 243 = 81 381 is a perfect square and a factor of 243. = 81 3Use the Multiplication Property of.

Solving Radical EquationsSolving Radical Equations

Solve each radical equation.

1. 7x – 3 = 4 2. 3x – 2 = x + 2

3. 2x + 7 = 5x – 8 4. x = 2x + 8

5. 3x + 4 + 5 = 3

2

5 4

no solution

2 57

ALGEBRA 1 LESSON 11-3ALGEBRA 1 LESSON 11-3

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