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UNIT 6.5 SOLVING SQUARE UNIT 6.5 SOLVING SQUARE ROOT AND OTHER RADICAL ROOT AND OTHER RADICAL EXPRESSIONS EXPRESSIONS

Algebra 2 unit 6.5

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Page 1: Algebra 2 unit 6.5

UNIT 6.5 SOLVING SQUARE UNIT 6.5 SOLVING SQUARE ROOT AND OTHER RADICALROOT AND OTHER RADICAL

EXPRESSIONSEXPRESSIONS

Page 2: Algebra 2 unit 6.5

Warm Up

Solve each equation.

1. 3x +5 = 17

2. 4x + 1 = 2x – 3

3.

4. (x + 7)(x – 4) = 0

5. x2 – 11x + 30 = 0

6. x2 = 2x + 15

4

–2

35

–7, 4

6, 5

5, –3

Page 3: Algebra 2 unit 6.5

Solve radical equations.

Objective

Page 4: Algebra 2 unit 6.5

radical equationextraneous solution

Vocabulary

Page 5: Algebra 2 unit 6.5

A radical equation is an equation that contains a variable within a radical. In this course, you will only study radical equations that contain square roots.

Recall that you use inverse operations to solve equations. For nonnegative numbers, squaring and taking the square root are inverse operations. When an equation contains a variable within a square root, square both sides of the equation to solve.

Page 6: Algebra 2 unit 6.5
Page 7: Algebra 2 unit 6.5

Example 1A: Solving Simple Radical Equations

Solve the equation. Check your answer.

x = 25

Square both sides.

Substitute 25 for x in the original equation.

55

5

Simplify.

Check

Page 8: Algebra 2 unit 6.5

Example 1B: Solving Simple Radical Equations

Solve the equation. Check your answer.

100 = 2x50 = x

Square both sides.

Divide both sides by 2.

Check

10 10

Substitute 50 for x in the original equation.

Simplify.

Page 9: Algebra 2 unit 6.5

Check It Out! Example 1a

Solve the equation. Check your answer.

Square both sides.

6 6

Check

Substitute 36 for x in the original equation.

Simplify.

Simplify.

Page 10: Algebra 2 unit 6.5

Check It Out! Example 1b

Solve the equation. Check your answer.

81 = 27x3 = x

Square both sides.

Divide both sides by 27.

Substitute 3 for x in the original equation.

Simplify.

Check

Page 11: Algebra 2 unit 6.5

Check It Out! Example 1c Solve the equation. Check your answer.

3x = 1

Check

Square both sides.

Divide both sides by 3.

Simplify.

Substitute for x in the original equation.

Page 12: Algebra 2 unit 6.5

Some square-root equations do not have the square root isolated. To solve these equations, you may have to isolate the square root before squaring both sides. You can do this by using one or more inverse operations.

Page 13: Algebra 2 unit 6.5

Example 2A: Solving Simple Radical Equations

Solve the equation. Check your answer.

x = 81

Add 4 to both sides.

Square both sides.

Check

9 – 4 55 5

Page 14: Algebra 2 unit 6.5

Example 2B: Solving Simple Radical Equations

Solve the equation. Check your answer.

x = 46 Subtract 3 from both sides.

Square both sides.

Check

7 7

Page 15: Algebra 2 unit 6.5

Example 2C: Solving Simple Radical Equations

Solve the equation. Check your answer.

5x + 1 = 16

5x = 15

x = 3

Subtract 6 from both sides.

Square both sides.

Subtract 1 from both sides.

Divide both sides by 5.

Page 16: Algebra 2 unit 6.5

Example 2C Continued

Solve the equation. Check your answer.

4 + 6 1010 10

Check

Page 17: Algebra 2 unit 6.5

Check It Out! Example 2a

Solve the equation. Check your answer.

x = 9

Add 2 to both sides.

Square both sides.

Check

1 1

Page 18: Algebra 2 unit 6.5

Check It Out! Example 2b

Solve the equation. Check your answer.

x = 18Subtract 7 from both sides.

Square both sides.

Check

5 5

Page 19: Algebra 2 unit 6.5

Check It Out! Example 2c

Solve the equation. Check your answer.

3x = 9

x = 3

Add 1 to both sides.

Square both sides.

Subtract 7 from both sides.

Divide both sides by 3.

Page 20: Algebra 2 unit 6.5

Check It Out! Example 2c

Solve the equation. Check your answer.

3 3

Check

Page 21: Algebra 2 unit 6.5

Example 3A: Solving Radical Equations by Multiplying or Dividing

Solve the equation. Check your answer.

Method 1

x = 64

Divide both sides by 4.

Square both sides.

Page 22: Algebra 2 unit 6.5

Example 3A Continued

Solve the equation. Check your answer.

Method 2

x = 64

Square both sides.

Divide both sides by 16.

Page 23: Algebra 2 unit 6.5

Example 3A Continued

Solve the equation. Check your answer.

32 32

Check

Substitute 64 for x in the original equation.

Simplify.

Page 24: Algebra 2 unit 6.5

Example 3B: Solving Radical Equations by Multiplying or Dividing

Solve the equation. Check your answer.

Method 1

144 = x

Square both sides.

Multiply both sides by 2.

Page 25: Algebra 2 unit 6.5

Example 3B Continued

Solve the equation. Check your answer.

Method 2

Square both sides.

Multiply both sides by 4.

144 = x

Page 26: Algebra 2 unit 6.5

Example 3B Continued

Solve the equation. Check your answer.

6 6

Check

Substitute 144 for x in the original equation.

Simplify.

Page 27: Algebra 2 unit 6.5

Check It Out! Example 3a

Solve the equation. Check your answer.

Method 1

Square both sides.

Divide both sides by 2.

Page 28: Algebra 2 unit 6.5

Check It Out! Example 3a Continued

Solve the equation. Check your answer.

Method 2

Square both sides.

Divide both sides by 4.

x = 121

Page 29: Algebra 2 unit 6.5

Check It Out! Example 3a Continued

Solve the equation. Check your answer.

Substitute 121 for x in the original equation.

Simplify.

Check

Page 30: Algebra 2 unit 6.5

Check It Out! Example 3b

Solve the equation. Check your answer.

Method 1

Square both sides.

Multiply both sides by 4.

64 = x

Page 31: Algebra 2 unit 6.5

Check It Out! Example 3b Continued

Solve the equation. Check your answer.

Method 2

Square both sides.

Multiply both sides by 16.

Page 32: Algebra 2 unit 6.5

Check It Out! Example 3b Continued

Solve the equation. Check your answer.

Substitute 64 for x in the original equation.

Simplify.

Check

Page 33: Algebra 2 unit 6.5

Check It Out! Example 3c

Solve the equation. Check your answer.

Method 1

Square both sides.

Multiply both sides by 5.

x = 100Divide both sides by 4.

Page 34: Algebra 2 unit 6.5

Check It Out! Example 3c Continued

Solve the equation. Check your answer.

Method 2

Square both sides.

Multiply both sides by 25.

4x = 400

x = 100 Divide both sides by 4.

Page 35: Algebra 2 unit 6.5

Check It Out! Example 3c Continued

Solve the equation. Check your answer.

Substitute 100 for x in the original equation.

Simplify.

Check

4 4

4

Page 36: Algebra 2 unit 6.5

Example 4A: Solving Radical Equations with Square Roots on Both Sides

Solve the equation. Check your answer.

2x – 1 = x + 7x = 8

Square both sides.

Add 1 to both sides and subtract x from both sides.

Check

Page 37: Algebra 2 unit 6.5

Example 4B: Solving Radical Equations with Square Roots on Both Sides

Solve the equation. Check your answer.

5x – 4 = 65x = 10x = 2

Add to both sides.

Square both sides.

Add 4 to both sides.

Divide both sides by 2.

Page 38: Algebra 2 unit 6.5

Example 4B Continued

Solve the equation. Check your answer.

Check

0 0

Page 39: Algebra 2 unit 6.5

Check It Out! Example 4a

Solve the equation. Check your answer.

2x = 4

x = 2

Square both sides.

Subtract x from both sides and subtract 2 from both sides.

Divide both sides by 2.

Page 40: Algebra 2 unit 6.5

Check It Out! Example 4a Continued

Solve the equation. Check your answer.

Check

Page 41: Algebra 2 unit 6.5

Check It Out! Example 4b

Solve the equation. Check your answer.

2x – 5 = 6

2x = 11

Add to both sides.

Square both sides.

Add 5 to both sides.

Divide both sides by 2.

Page 42: Algebra 2 unit 6.5

Check It Out! Example 4b Continued

Solve the equation. Check your answer.

Check

0 0

Page 43: Algebra 2 unit 6.5

Squaring both sides of an equation may result in an extraneous solution—a number that is not a solution of the original equation.

Suppose your original equation is x = 3.

Square both sides. Now you have a new equation.

Solve this new equation for x by taking the square root of both sides.

x = 3

x2 = 9

x = 3 or x = –3

Page 44: Algebra 2 unit 6.5

Now there are two solutions. One (x = 3) is the original equation. The other (x = –3) is extraneous–it is not a solution of the original equation. Because of extraneous solutions, it is important to check your answers.

Page 45: Algebra 2 unit 6.5

Example 5A: Extraneous Solutions

Solve Check your answer.

Square both sides

Divide both sides by 6.

Subtract 12 from each sides.

6x = 36

x = 6

Page 46: Algebra 2 unit 6.5

Example 5A Continued

Solve Check your answer.

Substitute 6 for x in the equation.

Check

6 does not check. There is no solution.

18 6

Page 47: Algebra 2 unit 6.5

Example 5B: Extraneous Solutions

Solve Check your answer.

x2 – 2x – 3 = 0

(x – 3)(x + 1) = 0

x – 3 = 0 or x + 1 = 0

x = 3 or x = –1

Square both sides

Write in standard form.

Factor.

Zero-Product Property

Solve for x.

x2 = 2x + 3

Page 48: Algebra 2 unit 6.5

Example 5B Continued

Solve Check your answer.

Substitute –1 for x in the equation.

Check

–1 1

Substitute 3 for x in the equation.

3 3–1 does not check; it is extraneous. The only solution is 3.

Page 49: Algebra 2 unit 6.5

Check It Out! Example 5a

Solve the equation. Check your answer.

x = 5

Subtract 11 from both sides.

Square both sides.

Simplify.

Page 50: Algebra 2 unit 6.5

Check It Out! Example 5a Continued

Solve the equation. Check your answer.

Substitute 5 for x in the equation.

16 6

Check

No solution. The answer is extraneous.

Page 51: Algebra 2 unit 6.5

Check It Out! Example 5b

Solve the equation. Check your answer.

x2 = –3x – 2

x2 + 3x + 2 = 0

(x + 1)(x + 2) = 0

x = –1 or x = –2

Square both sides

Write in standard form.

Factor.

Zero-Product Property

Solve for x.

x + 1 = 0 or x + 2 = 0

Page 52: Algebra 2 unit 6.5

Check It Out! Example 5b Continued

Solve the equation. Check your answer.

Substitute –1 for x in the equation.

Check

–2 2Substitute –2 for x in the

equation.

No solutions. Both answers are extraneous.

Page 53: Algebra 2 unit 6.5

Check It Out! Example 5c

Solve the equation. Check your answer.

x2 – 5x +4 = 0

(x – 1)(x – 4) = 0

x = 1 or x = 4

X – 1 = 0 or x – 4 = 0

Square both sides

Write in standard form.

Factor.

Zero-Product Property.

Solve for x.

Page 54: Algebra 2 unit 6.5

Check It Out! Example 5c Continued

Solve the equation. Check your answer.

Substitute 1 for x in the equation.

Substitute 4 for x in the equation.

1 does not check; it is extraneous. The only solution is 4.

Check

2 2

Page 55: Algebra 2 unit 6.5

Example 6: Geometry Application

8 ft

Use the formula for area of a triangle.

Substitute 8 for b, 36 for A and for h.

Divide both sides by 4.

A triangle has an area of 36 square feet, its base is 8 feet, and its height is feet. What is the value of x? What is the height of the triangle?

Simplify.

Page 56: Algebra 2 unit 6.5

Example 6 Continued

8 ft

82 = x

Square both sides.

A triangle has an area of 36 square feet, its base is 8 feet, and its height is feet. What is the value of x? What is the height of the triangle?

81 = x – 1

Page 57: Algebra 2 unit 6.5

8 ft

Check

36 36

The value of x is 82. The height of the triangle is 9 feet.

Substitute 82 for x.

Example 6 ContinuedA triangle has an area of 36 square feet, its base is 8 feet, and its height is feet. What is the value of x? What is the height of the triangle?

Page 58: Algebra 2 unit 6.5

Check It Out! Example 6

A rectangle has an area of 15 cm2. Its width is 5 cm, and its length is ( ) cm. What is the value of x? What is the length of the rectangle?

5

A = lw Use the formula for area of a rectangle.

Divide both sides by 5.

Substitute 5 for w, 15 for A, and for l.

Page 59: Algebra 2 unit 6.5

Check It Out! Example 6 Continued

A rectangle has an area of 15 cm2. Its width is 5 cm, and its length is ( ) cm. What is the value of x? What is the length of the rectangle?

5

8 = x

Square both sides.

The value of x is 8. The length of the rectangle is cm.

Page 60: Algebra 2 unit 6.5

Check A = lw

15 15

Substitute 8 for x.

Check It Out! Example 6 Continued

A rectangle has an area of 15 cm2. Its width is 5 cm, and its length is ( ) cm. What is the value of x? What is the length of the rectangle?

5

Page 61: Algebra 2 unit 6.5

Lesson Quiz: Part I

Solve each equation. Check your answer.

1.

3.

5.

2.

4.

6.

36 45

no solution 11

4 4

Page 62: Algebra 2 unit 6.5

Lesson Quiz: Part II

7. A triangle has an area of 48 square feet, its base is 6 feet and its height is feet. What is the value of x? What is the height of the triangle?

253; 16 ft

Page 63: Algebra 2 unit 6.5

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