11-9 Solving Radical Equations

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


1.  Students will be able to solve radical equations 2.  Students will be able to solve one-step radical

equations using addition and subtraction 3.  Students will be able to solve multi-step

radical equations using multiplication and division

4.  Students will be able to solve radical equations with multiple radicals

Learning Goal

11-9 Solving Radical Equations A radical equation is an equation that contains a variable within a radical.

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.

42 =16 42( ) = 16 42( ) = 4 4 = 4

25 = 5 25( )2= 5( )2 25= 25 52( ) = 5


11-9 Solving Radical Equations Example 1: Solving Simple Radical Equations Solve the equation. Check your answer.

10 = 2xB. x = 5A.


9 = 27xb. x = 6a.

20a =10e. x = 7d.

3x =1c.

9 = 27xb.


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.


x +3 = 7B. x − 4 = 5A. 5x +1+ 6 =10C.


x + 7 = 5b. x − 2 =1a. 3x + 7 −1= 3c.

2− a = 3e. x + 6 =11d.

11-9 Solving Radical Equations Example 3: Solving Radical Equations by Multiplying

or Dividing Solve the equation. Check your answer.

6 = x2

B. 4 x = 32A.


2 = x4

b. 2 x = 22a. 2 x5

= 4c.

Example 3: Solving Radical Equations by Multiplying or Dividing

Solve the equation. Check your answer.

2 x = 8e. 5 x6

=10d. x3= 3f.

3 x2

=1g. 13 2x = 26h. x5= 2i.

x − 73

=1j. 4 2x −1 =12k.

11-9 Solving Radical Equations Example 4: Solving Radical Equations with Square

Roots on Both Sides Solve the equation. Check your answer.

5x − 4 − 6 = 0B. 2x −1 = x + 7A.


2x − 5 = 6 = 0b. 3x + 2 = x + 6a.

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

Solve the equation. Check your answer.

0 = 2x − x +3d. 5− x = 6x − 2c.

x − 5 + 5= 0f. −x = 2x +1e.


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


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.

11-9 Solving Radical Equations Example 5A: Extraneous Solutions

Solve. Check your answer. A. B.

11-9 Solving Radical Equations Check It Out! Example 5a

Solve the equation. Check your answer. a. b. c.

11-9 Solving Radical Equations Example 6: Geometry Application

8 ft

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?


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


Lesson Quiz: Part I

Solve each equation. Check your answer.

1.

3.

5.

2.

4.

6.


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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11-9 Solving Radical Equations