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Algebra II – Chapter 6 Day #4
Topic: Solving Square Root and Other Radical Equations
Standards/Goals:
 G.1.g./A.REI.2: I can solve simple rational and radical equations in one variable.
o I can show how to arrive at ‘extraneous’ solutions.
 A.CED.4.: I can rearrange formulas to highlight a quantity of interest, using the same
techniques and methods as you would use to solve an equation.
We first want to focus on how to solve a square root equation.
Examples: What is the solution for the following equations?
#1. √𝑥 − 2 + 1 = 5
#2. √𝑥 + 4 + 6 = 7
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#3. (4𝑥 + 1)4 − 4 = −1
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#5. 3√(𝑥 + 1)3 + 1 = 25
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#4. (6𝑥 + 9)3 − 5 = −2
#6. √4𝑥 + 1 − 5 = 0
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We also want to introduce ourselves to the idea of an ‘extraneous solution.’
When you check your answers to an equation, a correct solution will give a true statement. An
extraneous solution will give a false statement.
Examples: What is the solution to the following equations? Check your results/answers.
#1. √𝑥 − 1 = 𝑥 − 7
#2. √5𝑥 + 14 = 𝑥
#3.
3x  2  2x  7  0
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HOMEWORK – Chapter 6 Day #4
Name ___________________________________ Date ______
Solve. To start, rewrite the equation to isolate the radical.
x2 20
2.
2x  3  7  0
3. 2  3x  2  6
4.
3x  2  7  0
1.
4x  3  2  5
5.
7.
3
2x  1  3
6.
8.
33  3x  3
2
3
2( x  2)  50
4
1
3
9. (6 x  5)  3  2
10.
1
2
11.
x 5  0
1
3
2x  2  0
1
2
12.
(7 x  3)  5
Solve. Check for extraneous solutions. To start, square each side of the equation.
13.
2x  7  x  2
14.
4x  5  x  2