Solve Equations Containing Square Roots
To solve an equation that contains a square root, we must first make sure
that the radical term is alone on one side of the equation and then clear
out the square root by squaring both sides of the equation. It is very
important to check the answer(s), as some answers actually are not
solutions. If an answer creates a no solution, that value is called
extraneous.
Example 1: Solve: √x = 5
The radical is alone one side, so square both sides
2
(√x) = (5)2
x = 25
Check the answer:
√25 = 5
5=5
True
Example 2: Solve: √x = −5
The radical is alone one side, so square both sides
2
(√x) = (−5)2
x = 25
Check the answer: √25 = −5
5 ≠ - 5 No Solution, Extraneous
Whenever we have a positive square root, we are finding the principal
root. The principal root is a positive number.
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 3: Solve:
√7x + 2 = 4
The radical is alone one side, so square both sides
2
(√7x + 2) = (4)2
7x + 2 = 16
Solve the resulting equation
-2 -2
Subtract 2 from both sides
7x
= 14 Divide both sides by 7
7
7
x=2
Check the answer: √7(2) + 2 = 4
√14 + 2 = 4
√16 = 4
4=4
Example 4: Solve:
True
−2√x − 3 + 9 = 3
The radical is not alone, so combine like terms before squaring
−2√x − 3 + 9 = 3
-9 -9
Subtract 9 from both sides
−2√x − 3
= −6
Divide both sides by -2
-2
-2
√x − 3 = 3
2
(√x − 3) = (3)2
x–3=9
+3+3
x = 12
Square both sides
Solve the resulting equation
Add 3 to both sides
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Check the answer: −2√12 − 3 + 9 = 3
−2√9 + 9 = 3
−2(3) + 9 = 3
−6 + 9 = 3
3=3
True
Example 5: Solve: √3x − 8 = √x
The radicals are alone on both sides, so square both sides
2
2
(√3x − 8) = (√x)
3x − 8 = x
Solve the resulting equation
- 3x
- 3x
Subtract 3x from both sides
- 8 = -2x Divide both sides by -2
-2 -2
4=x
Check the answer:
Example 6: Solve:
√3(4) − 8 = √4
√12 − 8 = 2
√4 = 2
2 = 2 True
√4x + 1 + 5 = x
The radical is not alone, so combine like terms before squaring
√4x + 1 + 5 = x
-5=-5
Subtract 5 from both sides
=x−5
√4x + 1
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
2
Square both sides
(√4x + 1) = (x − 5)2
4x + 1 = x 2 − 10x + 25 Multiply the square binomial
- 4x - 1
- 4x - 1 Subtract 4x and 1 from both sides
2
0 = x – 14x + 24
0 = (x – 12)(x – 2)
Factor the polynomial
x – 12 = 0 and x – 2 = 0
Set each factor equal to zero
+ 12 +12
+2+2
Solve each equation
x = 12
x=2
Check each answer:
√4(12) + 1 + 5 = 12
√48 + 1 + 5 = 12
√49 + 5 = 12
7 + 5 = 12
12 = 12
√4(2) + 1 + 5 = 2
True
√8 + 1 + 5 = 2
√9 + 5 = 2
3 +5=2
8 = 2 Not True
Extraneous Answer
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)