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Solving Square Root Equations

Solve radical equations and reject extraneous candidates. Includes a worked example, printable practice, and an answer key with explanations.

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Grade 11 Math AlgebraSolving Square Root Equations

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Includes

2 Text2 Short AnswerTrue / FalseFill in the Blanks

Standards

CCSS.MATH.CONTENT.HSA.REI.A.2

Topics

AlgebraSquare Root EquationsExtraneous SolutionsGrade 11
8 sections · Free to use · Printable
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Solving Square Root Equations

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Learning goal and worked example

Solve radical equations and reject extraneous candidates.

For √(x+2)=x, require x≥0. Squaring gives x²−x−2=0, so candidates are 2 and −1. Only 2 satisfies the original equation.

The principal square root is nonnegative. Isolate it, square, solve, and substitute into the original equation.

Solve each square root equation. Be sure to check your solutions for extraneous roots.

1. Solve for x: √(x + 5) = 3

2. Find the solution(s) to: √(2x - 1) + 4 = 7

3. Determine the value of y in: 5 - √(y + 2) = 2

Indicate whether the statement is True or False.

4. The equation √(x) = -2 has a real solution.

T

True

F

False

5. When solving √(x + 1) = x - 1, squaring both sides always leads to the correct solution(s) without introducing extraneous ones.

T

True

F

False

Fill in the blank with the correct term.

6. An   solution is a solution that arises from the process of solving an equation but is not a valid solution to the original equation.

7. To check for extraneous solutions in square root equations, you must substitute your answer(s) back into the   equation.

8. Solve the equation √(x - 3) = x - 5. Clearly show all steps and identify any extraneous solutions.

9. Consider the equation √(3x + 4) = x. What are the potential solutions? Which of them, if any, are extraneous?

Answer key and worked solutions

1. x+5=9, so x=4; check √9=3.

2. √(2x−1)=3, so 2x−1=9 and x=5; check 3+4=7.

3. √(y+2)=3, so y=7; check 5−3=2.

4. False: a principal square root cannot equal −2.

5. False: squaring yields candidates x=0 and 3, but x=0 gives 1=−1 in the original equation; only x=3 works.

6. Extraneous. 7. Original.

8. Require x≥5. Squaring gives x−3=x²−10x+25, or (x−7)(x−4)=0. Candidate 7 works (2=2); candidate 4 is extraneous (1≠−1). Solution: x=7.

9. Require x≥0. Squaring gives x²−3x−4=(x−4)(x+1)=0. Candidate 4 works (4=4); −1 is extraneous (1≠−1). Solution: x=4.

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