Holt Algebra 1 11-9 Solving Radical Equations 11-9 Solving Radical Equations Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson

Holt Algebra 1

11-9 Solving Radical Equations11-9 Solving Radical Equations

Holt Algebra 1

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Algebra 1

11-9 Solving Radical Equations

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

Holt Algebra 1


Solve radical equations.

Objective

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radical equationextraneous solution

Vocabulary

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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.

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Holt Algebra 1


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

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Example 1B: Solving Simple Radical Equations


100 = 2x

50 = x

Square both sides.

Divide both sides by 2.

Check

10 10


Simplify.

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Check It Out! Example 1a


Square both sides.

6 6

Check


Simplify.

Simplify.

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Check It Out! Example 1b


81 = 27x

3 = x

Square both sides.



Simplify.

Check

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Check It Out! Example 1c


3x = 1

Check

Square both sides.


Simplify.

Substitute for x in the original equation.

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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.

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Example 2A: Solving Simple Radical Equations


x = 81

Add 4 to both sides.

Square both sides.

Check

9 – 4 55 5

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Example 2B: Solving Simple Radical Equations


x = 46 Subtract 3 from both sides.

Square both sides.

Check

7 7

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Example 2C: Solving Simple Radical Equations


5x + 1 = 16

5x = 15

x = 3

Subtract 6 from both sides.

Square both sides.



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Example 2C Continued


4 + 6 10

10 10

Check

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x = 9


Square both sides.

Check

1 1

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x = 18


Square both sides.

Check

5 5

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3x = 9

x = 3


Square both sides.



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3 3

Check

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Example 3A: Solving Radical Equations by Multiplying or Dividing


Method 1

x = 64


Square both sides.

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Example 3A Continued


Method 2

x = 64

Square both sides.


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32 32

Check


Simplify.

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Example 3B: Solving Radical Equations by Multiplying or Dividing


Method 1

144 = x

Square both sides.

Multiply both sides by 2.

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Example 3B Continued


Method 2

Square both sides.


144 = x

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6 6

Check


Simplify.

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Method 1

Square both sides.


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Check It Out! Example 3a Continued


Method 2

Square both sides.


x = 121

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Simplify.

Check

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Method 1

Square both sides.


64 = x

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Check It Out! Example 3b Continued


Method 2

Square both sides.


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Simplify.

Check

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Method 1

Square both sides.


x = 100Divide both sides by 4.

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Check It Out! Example 3c Continued


Method 2

Square both sides.


4x = 400

x = 100 Divide both sides by 4.

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Simplify.

Check

4 4

4

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


2x – 1 = x + 7

x = 8

Square both sides.

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

Check

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


5x – 4 = 6

5x = 10

x = 2

Add to both sides.

Square both sides.



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Check

0 0

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2x = 4

x = 2

Square both sides.

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


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Check

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2x – 5 = 6

2x = 11

Add to both sides.

Square both sides.



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Check

0 0

Holt Algebra 1


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

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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.

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Example 5A: Extraneous Solutions

Solve Check your answer.

Square both sides


Subtract 12 from each sides.

6x = 36

x = 6

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Substitute 6 for x in the equation.

Check

6 does not check. There is no solution.

18 6

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Example 5B: Extraneous Solutions


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

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Substitute –1 for x in the equation.

Check

–1 1


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

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x = 5


Square both sides.

Simplify.

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16 6

Check

No solution. The answer is extraneous.

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x2 = –3x – 2

x2 + 3x + 2 = 0

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

x = –1 or x = –2

Square both sides


Factor.

Zero-Product Property

Solve for x.

x + 1 = 0 or x + 2 = 0

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Substitute –1 for x in the equation.

Check

–2 2Substitute –2 for x in the

equation.

No solutions. Both answers are extraneous.

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x2 – 5x +4 = 0

(x – 1)(x – 4) = 0

x = 1 or x = 4

X – 1 = 0 or x – 4 = 0

Square both sides


Factor.

Zero-Product Property.

Solve for x.

Holt Algebra 1






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

Check

2 2

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Example 6: Geometry Application

8 ft

Use the formula for area of a triangle.

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


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.

Holt Algebra 1


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

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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?

Holt Algebra 1


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.


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

Holt Algebra 1


Check It Out! Example 6 Continued


5

8 = x

Square both sides.

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

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Check A = lw

15 15

Substitute 8 for x.

Check It Out! Example 6 Continued


5

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Lesson Quiz: Part I

Solve each equation. Check your answer.

1.

3.

5.

2.

4.

6.

36 45

no solution 11

4 4

Holt Algebra 1


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

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Holt Algebra 1 11-9 Solving Radical Equations 11-9 Solving Radical Equations Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson