Square root equations: solve, then catch the extraneous solution

By Marius, private math tutor in Montreal · 2026-10-05

To solve a square root equation, isolate the square root, square both sides, solve the equation that results, and then substitute every answer into the original equation: an answer that fails is an extraneous solution, and you reject it. It appears because squaring cannot tell $\sqrt{A} = B$ apart from $\sqrt{A} = - B$ , so you get the solutions of both equations at once.

The method

Isolate the square root on one side of the equation, and write down the domain: the expression under the root cannot be negative.
Square both sides, squaring each whole side: $(2 x - 3)^{2}$ is $4 x^{2} - 12 x + 9$ , not just the squares of the two terms.
Solve the new equation. When $x$ also appears outside the root, it is usually a second degree equation with two candidates.
Substitute each candidate into the original equation. Keep the ones that work. The ones that fail are extraneous.

Worked example 1: one answer survives

Solve $\sqrt{2 x + 3} + 3 = 2 x$ .

Domain: $2 x + 3 \geq 0$ , so $x \geq - \frac{3}{2}$ . Isolate the root by subtracting $3$ :

\sqrt{2 x + 3} = 2 x - 3

Square both sides:

2 x + 3 = (2 x - 3)^{2}

2 x + 3 = 4 x^{2} - 12 x + 9

0 = 4 x^{2} - 14 x + 6

Divide by $2$ and factor:

0 = 2 x^{2} - 7 x + 3 = (2 x - 1) (x - 3)

The candidates are $x = \frac{1}{2}$ and $x = 3$ . Now substitute each one into the original equation:

$x = 3$ : the left side is $\sqrt{9} + 3 = 6$ and the right side is $2 (3) = 6$ . ✓ A solution.
$x = \frac{1}{2}$ : the left side is $\sqrt{4} + 3 = 5$ and the right side is $2 (\frac{1}{2}) = 1$ . ✗ Extraneous.

The only solution is $x = 3$ .

Where the fake solution comes from

Look at $x = \frac{1}{2}$ in the isolated equation $\sqrt{2 x + 3} = 2 x - 3$ . The left side is $\sqrt{4} = 2$ and the right side is $2 (\frac{1}{2}) - 3 = - 2$ . The two sides have the same size and opposite signs, and that is exactly the difference squaring erases: $2^{2}$ and $(- 2)^{2}$ are both $4$ .

So $x = \frac{1}{2}$ is not a random accident. It is a genuine solution of a different equation,

\sqrt{2 x + 3} = - (2 x - 3)

and that equation squares to exactly the same thing. Squaring merges the two equations into one, and the quadratic you solve hands you the solutions of both. The graph shows it.

Squaring turns

\sqrt{2 x + 3} = 2 x - 3

into

2 x + 3 = (2 x - 3)^{2}

, which holds wherever the line

y = 2 x - 3

meets either branch. The solid branch is

y = \sqrt{2 x + 3}

, the original equation; the dashed branch,

y = - \sqrt{2 x + 3}

, is what squaring adds. Only the crossing on the solid branch,

x = 3

, solves the original equation.

Spotting it before you substitute

A square root is never negative, so in $\sqrt{A} = B$ the side $B$ cannot be negative either. That condition is precisely what separates the equation from its twin $\sqrt{A} = - B$ : a candidate that makes $B$ negative belongs to the twin.

In Worked example 1, $B = 2 x - 3$ , and $2 x - 3 \geq 0$ means $x \geq \frac{3}{2}$ . The candidate $\frac{1}{2}$ is below that, so it can be rejected on sight. The domain $x \geq - \frac{3}{2}$ does not catch it: $\frac{1}{2}$ is inside the domain, and every candidate always is, because it makes $A$ equal to $B^{2}$ , which is never negative. Treat the condition on $B$ as a shortcut, not a replacement: substituting still catches arithmetic slips that the condition cannot see.

Worked example 2: both answers survive

Solve $\sqrt{5 x - 1} = x + 1$ .

Domain: $5 x - 1 \geq 0$ , so $x \geq \frac{1}{5}$ . The root is already alone. Square both sides:

5 x - 1 = x^{2} + 2 x + 1

0 = x^{2} - 3 x + 2 = (x - 1) (x - 2)

$x = 1$ : $\sqrt{4} = 2$ and $1 + 1 = 2$ . ✓
$x = 2$ : $\sqrt{9} = 3$ and $2 + 1 = 3$ . ✓

Both candidates are solutions, and the solution set is ${1, 2}$ . The habit to build is checking, not throwing one answer away. Squaring can add a solution; whether it actually did depends on the equation, and only the check says which.

The most common mistake: checking in the wrong equation. The check goes into the original equation, the one with the square root in it. Every candidate satisfies the squared equation automatically, because that is the equation it was solved from, so a check there waves the extraneous value through with the real one. In Worked example 1, $x = \frac{1}{2}$ makes $2 x + 3$ and $(2 x - 3)^{2}$ both equal to $4$ , and it still fails $\sqrt{2 x + 3} + 3 = 2 x$ .

Check your understanding

No calculator needed. Questions 4 to 7 ask for reasons as well as answers.

Solve $\sqrt{2 x - 1} = 3$ .
Solve $\sqrt{4 x + 1} = 2 x - 1$ .
Solve $\sqrt{x + 13} = 7 - x$ . Which candidate does the condition $7 - x \geq 0$ let you reject before you substitute?
True or false: in $\sqrt{x + 5} = x + 3$ , the candidate $x = - 1$ must be rejected because it is negative.
Find the error. To solve $\sqrt{2 x + 9} + 3 = x$ , Noah squares both sides straight away and writes $2 x + 9 + 9 = x^{2}$ .
Kim checks the rejected value $x = 0$ from question 2 in the squared equation $4 x + 1 = (2 x - 1)^{2}$ . Both sides equal $1$ , so she decides it is a solution after all. What is wrong with her check?
The equation $\sqrt{5 x - 1} = - x - 1$ squares to the same quadratic as Worked example 2. What is its solution set, and why?

Show the answers

$x = 5$ . The domain is $x \geq \frac{1}{2}$ . Squaring gives $2 x - 1 = 9$ , so $x = 5$ , and $\sqrt{9} = 3$ checks. The right side, $3$ , is positive, so squaring adds nothing extra here.
$x = 2$ . The domain is $x \geq - \frac{1}{4}$ . Squaring: $4 x + 1 = 4 x^{2} - 4 x + 1$ , so $4 x^{2} - 8 x = 0$ and $4 x (x - 2) = 0$ , giving $x = 0$ or $x = 2$ . At $x = 2$ : $\sqrt{9} = 3$ and $2 (2) - 1 = 3$ . At $x = 0$ : $\sqrt{1} = 1$ but $2 (0) - 1 = - 1$ , so $0$ is rejected.
$x = 3$ . The domain is $x \geq - 13$ . Squaring: $x + 13 = 49 - 14 x + x^{2}$ , so $x^{2} - 15 x + 36 = 0$ and $(x - 3) (x - 12) = 0$ . The condition $7 - x \geq 0$ means $x \leq 7$ , so $12$ is rejected on sight (it makes the right side $- 5$ ). At $x = 3$ : $\sqrt{16} = 4$ and $7 - 3 = 4$ .
False. $x = - 1$ is in the domain $x \geq - 5$ , and $\sqrt{- 1 + 5} = 2$ and $- 1 + 3 = 2$ , so it is a solution. What cannot be negative is the value of the square root, and so the side it equals, not $x$ itself. The other candidate, $x = - 4$ , is the extraneous one: $\sqrt{1} = 1$ but $- 4 + 3 = - 1$ .
The square of a sum is not the sum of the squares: $(\sqrt{2 x + 9} + 3)^{2}$ is $2 x + 18 + 6 \sqrt{2 x + 9}$ , so the root has not gone. Isolate it first: $\sqrt{2 x + 9} = x - 3$ , with domain $x \geq - \frac{9}{2}$ , then $2 x + 9 = x^{2} - 6 x + 9$ , so $x^{2} - 8 x = 0$ and $x = 0$ or $x = 8$ . At $x = 8$ the left side is $\sqrt{25} + 3 = 8$ , which checks. At $x = 0$ it is $\sqrt{9} + 3 = 6$ , not $0$ , so $0$ is rejected. The solution is $x = 8$ .
$x = 0$ came from the squared equation, so of course it satisfies it; that check cannot tell a real solution from an extraneous one. In the original, $\sqrt{1} = 1$ while $2 (0) - 1 = - 1$ , so $0$ is still rejected.
No solution: the solution set is $\emptyset$ . Squaring gives $5 x - 1 = x^{2} + 2 x + 1$ again, with candidates $1$ and $2$ , but now the right side is $- 2$ and $- 3$ at those values while the root is $2$ and $3$ . This is the twin of Worked example 2. Faster still: the root exists only for $x \geq \frac{1}{5}$ , and there $- x - 1$ is always negative.

If question 4 caught you, remember that the sign rule is about the value of the root, not about $x$ . If question 5 did, isolate before you square, every time.

Practise it

Square root equations have their own sheet on the free Secondary 4 Equations and Inequalities worksheet, beside the second degree, exponential and greatest integer ones. They come back on the Secondary 5 worksheet of the same name, and the graph of the square root function itself is on the Secondary 4 Square Root Functions worksheet.

Want one-on-one help with this? I offer Secondary 4 and 5 math tutoring in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.