When do I know if my solution is extraneous?

Substitute it back into the original equation. If both sides are equal, it's valid; if not, it's extraneous. An extraneous solution satisfies the squared equation but not the original radical equation.

Why do extraneous solutions appear when solving radical equations?

Squaring both sides can introduce solutions that don't satisfy the original equation. Since \sqrt{f(x)} is always non-negative, the equation \sqrt{f(x)} = g(x) requires g(x) \geq 0. Squaring removes this restriction, allowing invalid solutions to sneak in.

What should I do if I have a radical on both sides of the equation?

Square both sides just as usual. For example, \sqrt{x} = \sqrt{x + 1} becomes x = x + 1 after squaring, which has no solution. Always check your answers in the original equation.

What if I get no solutions after checking?

That's perfectly valid. For example, \sqrt{x} = -2 has no real solutions because a square root is never negative. After squaring, you'd get x = 4, but checking shows \sqrt{4} = 2 \neq -2. The extraneous solution must be discarded.

Radical Equations

Solve equations with square roots by squaring both sides to eliminate the radical, then check for extraneous solutions that don't satisfy the original equation.

\sqrt{f(x)} = g(x) \implies f(x) = g(x)^2

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What each symbol means

What Radical Equations takes

Symbol	Meaning
$f$	The expression under the square root sign; $f(x)$ can be a simple variable, a linear expression, a polynomial, or any function, but it must be non-negative in the real numbers for the square root to be defined.
$g$	The expression on the other side of the equation equal to the square root; $g(x)$ can be any expression, but since a square root is never negative, $g(x)$ must be non-negative for a real solution to exist.
$x$	The unknown variable you are solving for; it may appear in $f(x)$ , in $g(x)$ , or in both, and the goal is to find all values of $x$ that make the original equation true.

When to use it

When an unknown variable appears inside a square root on one side of an equation, square both sides to clear the radical and solve.

Level

Usually taught in: Algebra II

Worked examples

1. Solve a basic radical equation with a constant

Problem

Solve

\sqrt{x + 3} = 5

for

x

$\sqrt{x + 3} = 5$
This is the equation to solve.
$x + 3 = 25$
Square both sides to eliminate the square root.
$x = 22$
Subtract 3 from both sides to isolate $x$ .

Answer: $x = 22$

We squared both sides to eliminate the radical, then solved the resulting linear equation. To verify, substitute back: $\sqrt{22 + 3} = \sqrt{25} = 5$ ✓. This straightforward approach works well when the radical is already isolated on one side.

2. Solve a radical equation with a variable on the other side (extraneous solution)

Problem

Solve

\sqrt{2x - 1} = x - 2

for

x

$\sqrt{2x - 1} = x - 2$
This is our starting equation.
$2x - 1 = (x - 2)^2$
Square both sides to eliminate the radical.
$2x - 1 = x^2 - 4x + 4$
Expand the right side using $(a - b)^2 = a^2 - 2ab + b^2$ .
$0 = x^2 - 6x + 5$
Rearrange to standard quadratic form.
$0 = (x - 1)(x - 5)$
Factor the quadratic.
$x = 1 \text{ or } x = 5$
Apply the Zero Product Property to find both solutions.
$\sqrt{2(1) - 1} = 1 \text{ but } 1 - 2 = -1$
Check $x = 1$ : the left side equals 1, but the right side equals $-1$ , so $x = 1$ is extraneous.
$\sqrt{2(5) - 1} = 3 \text{ and } 5 - 2 = 3$
Check $x = 5$ : both sides equal 3, so $x = 5$ is a valid solution.

Answer: $x = 5$

This example demonstrates why checking is essential: squaring both sides produced an extraneous solution. We solved the resulting quadratic using factorization and the Zero Product Property, but $x = 1$ satisfies the squared equation while violating the original constraint that both sides must be equal. This is the critical lesson in radical equations: always verify your solutions.

3. Solve a word problem involving a radical equation

Problem

A school bake sale finds that the number of items sold (in dozens) follows the model

y = \sqrt{400 - x^2}

, where

x

is the price in dollars. What price should they set to sell exactly 12 dozen items?

$\sqrt{400 - x^2} = 12$
Set up the equation: we want $y = 12$ dozen items.
$400 - x^2 = 144$
Square both sides to eliminate the radical.
$-x^2 = -256$
Subtract 400 from both sides.
$x^2 = 256$
Multiply both sides by $-1$ .
$x = \pm 16$
Take the square root of both sides.
$x = 16$
Since price must be positive, discard $x = -16$ ; negative prices have no real-world meaning.

Answer: $x = 16 \text{ dollars}$

Real-world radical equations often appear in modeling situations. After solving the squared equation, we obtained two mathematical solutions, but only the positive one applies. We verified by substitution: $\sqrt{400 - 256} = \sqrt{144} = 12$ ✓. Always check whether your solution makes practical sense in context.

Common mistakes

Where Radical Equations usually goes wrong

The mistake	Why it is wrong	The fix
Squaring without isolating the radical first. For example, from $\sqrt{x + 1} + 2 = 5$ , writing $(x + 1) + 4 = 25$ .	If you square a binomial like $(\sqrt{x+1} + 2)^2$ , the binomial expansion $(a + b)^2 = a^2 + 2ab + b^2$ introduces cross terms that complicate the algebra and create unnecessary work.	Isolate the radical first: $\sqrt{x + 1} = 3$ . Then square: $x + 1 = 9$ , so $x = 8$ .
Forgetting to check solutions for extraneous answers. Solving $\sqrt{2x - 1} = x - 3$ and reporting both $x = 1$ and $x = 5$ without substituting back.	Squaring both sides can produce extraneous solutions that satisfy the squared equation but not the original. For $x = 1$ : $\sqrt{2(1)-1} = 1$ but $1 - 3 = -2$ , so the two sides are not equal.	Always substitute each solution back into the original equation. Only accept solutions where both sides equal the same number.
Incorrectly applying the square root to a sum. Writing $\sqrt{x + 4} = \sqrt{x} + \sqrt{4} = \sqrt{x} + 2$ .	The square root does not distribute over addition: $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$ in general. This violates fundamental properties of radicals.	Never split a square root across addition or subtraction. Keep the entire expression under the radical: $\sqrt{x+4}$ . Then proceed: $\sqrt{x+4} = 6 \Rightarrow x + 4 = 36 \Rightarrow x = 32$ .

Tips and when to use something else

Isolate the radical term on one side of the equation before squaring both sides; this keeps the algebra simpler and reduces errors.
Always substitute your final answers back into the original equation to verify them—squaring can introduce extraneous solutions that look correct but don't actually work.
For equations with multiple radicals, isolate one radical, square, and repeat to eliminate them one at a time rather than trying to square everything at once.
If the equation becomes a complicated polynomial (degree 3 or higher) after squaring, consider whether Solving by Substitution could simplify the problem first.

Related formulas

Extraneous SolutionsA solution to the transformed equation that doesn't satisfy the original; found when squaring both sides or applying other non-reversible operations.Square RootSquare root finds the non-negative number that, when multiplied by itself, gives your starting number; use it to undo squaring in equations.Solving Quadratics by FactoringSolve quadratic equations by factoring the polynomial into binomials, then apply the Zero Product Property to find the solutions.Rational EquationsA rational equation has a polynomial fraction equal to zero; solve by setting the numerator to zero, then verify the denominator is nonzero.

Frequently asked questions

When do I know if my solution is extraneous?: Substitute it back into the original equation. If both sides are equal, it's valid; if not, it's extraneous. An extraneous solution satisfies the squared equation but not the original radical equation.
Why do extraneous solutions appear when solving radical equations?: Squaring both sides can introduce solutions that don't satisfy the original equation. Since $\sqrt{f(x)}$ is always non-negative, the equation $\sqrt{f(x)} = g(x)$ requires $g(x) \geq 0$ . Squaring removes this restriction, allowing invalid solutions to sneak in.
What should I do if I have a radical on both sides of the equation?: Square both sides just as usual. For example, $\sqrt{x} = \sqrt{x + 1}$ becomes $x = x + 1$ after squaring, which has no solution. Always check your answers in the original equation.
What if I get no solutions after checking?: That's perfectly valid. For example, $\sqrt{x} = -2$ has no real solutions because a square root is never negative. After squaring, you'd get $x = 4$ , but checking shows $\sqrt{4} = 2 \neq -2$ . The extraneous solution must be discarded.

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Reviewed 2026-09-18