Georgia Performance Standard Essential Question

• MM1A3 Students will solve simple equations.• b. Solve equations involving radicals such as

√x + b = c using algebraic techniques.

• Essential Question: How do you solve equations including radicals such as √x + b = c?

Activating StrategySolve each equation.1. 3x +5 = 172. 4x + 1 = 2x – 3 3. 4. (x + 7)(x – 4) = 05. x2 – 11x + 30 = 06. x2 = 2x + 15

4–2

35–7, 4 6, 5

5, –3

Solve radical equations.Objective

radical equationextraneous solution

Vocabulary

Example 1A: Solving Simple Radical Equations

Solve the equation. Check your answer.

x = 25Square both sides.

Substitute 25 for x in the original equation.

55 5

Simplify.

Check

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.

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

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

Example 2C: Solving Simple Radical Equations

Solve the equation. Check your answer.

5x + 1 = 165x = 15

x = 3

Subtract 6 from both sides.

Square both sides.

Subtract 1 from both sides.

Divide both sides by 5.

Example 3A: Solving Radical Equations by Multiplying or Dividing

Solve the equation. Check your answer.

x = 64

Divide both sides by 4.

Square both sides.

Example 3B: Solving Radical Equations by Multiplying or Dividing

Solve the equation. Check your answer.

144 = x

Square both sides.

Multiply both sides by 2.

Check It Out! Example 3c Solve the equation. Check your answer.

Square both sides.

Multiply both sides by 5.

x = 100Divide both sides by 4.

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

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.

Example 5A: Extraneous Solutions

Solve Check your answer.

Square both sides

Divide both sides by 6.

Subtract 12 from each sides.

6x = 36x = 6

Example 5A ContinuedSolve Check your answer.

Substitute 6 for x in the equation.

Check

6 does not check. There is no solution.

18 6

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.

Check It Out! Example 5c ContinuedSolve 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

Key Points to Remember• Always make sure the radical is isolated.• Use inverse operations to isolate the radical.• Make sure to square both sides to get rid of

the radical.• Remember to always check your work for

extraneous solutions.

Lesson Quiz (TOD) Solve each equation. Check your answer.

1.

3.

2.

4.

36 45

no solution 11