You are solving an equation and you see a radical sitting by itself on one side — something like \sqrt{2x + 3} = 5, or \sqrt{x + 7} = x - 1. Your instinct is right: square both sides and the radical disappears. But there is a catch every textbook warns about and every student forgets at least once: squaring can manufacture fake solutions. The equation you end up with can have roots the original never had, and if you do not hunt them down and reject them, your "answer" is wrong.

This page is about spotting a radical equation and the discipline that must travel with the technique.

The technique — square both sides

The moment you see

\sqrt{A} = B

where A is some expression (usually involving x) and B is another expression (a number, or something in x), the move is immediate:

A = B^2

Square both sides. The radical is gone. What is left is a polynomial equation — linear, quadratic, sometimes cubic — and you already know how to solve those. Isolate, factorise, quadratic formula, whatever the polynomial calls for.

So the recognition is simple: radical alone on one side \Rightarrow square both sides. If the radical is not alone — if it is tangled up with other terms on the same side — isolate it first, then square.

Why extraneous solutions appear

Squaring is not a reversible operation in the way adding or multiplying are. Here is why.

The equation \sqrt{A} = B carries an invisible piece of information: the left side \sqrt{A} is always non-negative, because the radical sign returns the principal root. So the equation is secretly saying "A = B^2 and B \geq 0."

When you square both sides, you get A = B^2 — but the constraint B \geq 0 is gone. The squared equation does not care whether B was positive or negative, because (-B)^2 = B^2. So the squared equation has more permissiveness than the original; it accepts values of B that the original rejected.

Any solution of the squared equation for which B turns out to be negative is called extraneous. It satisfies the squared equation but not the original.

This is not a bug in the method. It is a predictable consequence of the algebra. The fix is to check every candidate.

Worked example 1 — a clean case

Solve \sqrt{2x + 3} = 5.

Square both sides:

2x + 3 = 25

Solve:

2x = 22, \quad x = 11

Check in the original: \sqrt{2 \cdot 11 + 3} = \sqrt{25} = 5. Matches. Answer: x = 11.

Here the right-hand side was the constant 5, which is positive regardless of x, so no extraneous solution could appear. The check was a formality. In the next example it is not.

Worked example 2 — an extraneous solution appears

Solve \sqrt{x + 7} = x - 1.

Square both sides:

x + 7 = (x - 1)^2 = x^2 - 2x + 1

Rearrange:

x^2 - 3x - 6 = 0

Quadratic formula:

x = \frac{3 \pm \sqrt{9 + 24}}{2} = \frac{3 \pm \sqrt{33}}{2}

So x_1 = \dfrac{3 + \sqrt{33}}{2} \approx 4.37 and x_2 = \dfrac{3 - \sqrt{33}}{2} \approx -1.37.

Check x_1 \approx 4.37. LHS: \sqrt{4.37 + 7} = \sqrt{11.37} \approx 3.37. RHS: 4.37 - 1 = 3.37. Both sides agree, both positive. Accept.

Check x_2 \approx -1.37. LHS: \sqrt{-1.37 + 7} = \sqrt{5.63} \approx 2.37. RHS: -1.37 - 1 = -2.37. The LHS is positive, the RHS is negative. They disagree. Reject.

Final answer: x = \dfrac{3 + \sqrt{33}}{2}.

Notice what happened. The squared equation x^2 - 3x - 6 = 0 has two perfectly valid roots — the quadratic formula handed them over without complaint. But only one of them came from the original radical equation; the other is an artefact of the squaring step. Without the check, you would have reported both and been half-wrong.

Worked example 3 — clean integer case with extraneous

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

Square:

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

Rearrange:

x^2 - 7x = 0, \quad x(x - 7) = 0

So x = 0 or x = 7.

Check x = 0. LHS: \sqrt{1} = 1. RHS: 0 - 1 = -1. Mismatch. Reject.

Check x = 7. LHS: \sqrt{36} = 6. RHS: 7 - 1 = 6. Match. Accept.

Final answer: x = 7.

Again: two roots from the squared equation, only one survives the check.

Why the check is not optional

It is easy to stop at the squared equation. Two neat answers, algebra clean — your work looks complete. It is not.

An unchecked extraneous solution is a wrong answer on your paper, and examiners know this. JEE, CBSE, and ICSE questions involving radical equations are designed so that at least one candidate is extraneous; the question is testing whether you caught it. Writing "x = (3 \pm \sqrt{33})/2" when the answer is "x = (3 + \sqrt{33})/2 only" loses marks.

The check is the last step, but it is not optional decoration. It is the step that separates the right answer from the wrong one.

Two radicals — square twice

Solve \sqrt{x + 5} = \sqrt{2x - 3} + 1.

One radical is already isolated on the left. Square both sides:

x + 5 = (\sqrt{2x - 3} + 1)^2 = (2x - 3) + 2\sqrt{2x - 3} + 1

Simplify the non-radical terms on the right: (2x - 3) + 1 = 2x - 2. So:

x + 5 = 2x - 2 + 2\sqrt{2x - 3}

Isolate the remaining radical:

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

Square again:

(7 - x)^2 = 4(2x - 3)
49 - 14x + x^2 = 8x - 12

Rearrange:

x^2 - 22x + 61 = 0

Quadratic formula:

x = \frac{22 \pm \sqrt{484 - 244}}{2} = \frac{22 \pm \sqrt{240}}{2} = 11 \pm 2\sqrt{15}

Two candidates, x \approx 18.75 and x \approx 3.25. You must plug each back into the original equation and check — not into the intermediate squared forms, because those carry the extraneous roots. Typically one of the two will fail; work it out for yourself as practice.

The pattern for two radicals: isolate one, square, isolate the remaining one, square again. Check at the end.

The domain constraint — do not forget it

There is one more filter that can catch bad candidates even before the squaring-check: the domain. The expression under a radical must be non-negative.

For \sqrt{2x + 3} = 5, the domain is 2x + 3 \geq 0, i.e., x \geq -3/2. Any candidate below that is out immediately, without even plugging into the equation.

For \sqrt{x + 7} = x - 1, the domain is x + 7 \geq 0, i.e., x \geq -7. Both x_1 \approx 4.37 and x_2 \approx -1.37 pass this filter, which is why the final check had to be done by substitution — the domain was not restrictive enough here to do the rejecting by itself.

Use the domain as an early filter when it is tight; fall back to the full substitution check when it is not.

The habit

For every radical equation, walk through these five steps, every time:

  1. Isolate the radical on one side (if it is not already).
  2. Square both sides.
  3. Solve the resulting polynomial equation.
  4. Check every candidate in the original equation.
  5. Report only the candidates that passed the check.

Step 4 is the one students skip. Make it automatic.

Why extraneous solutions are often forgotten

Because step 3 is satisfying. The quadratic formula gave clean roots, the arithmetic worked, you wrote down two answers. Your brain marks the problem as "done" and moves on. The check at step 4 feels like a formality.

It is not. It is the step that detects the fake solution that squaring quietly inserted. Skipping it is submitting a wrong answer.

The complete recipe

Squaring is the correct move for radical equations. Extraneous solutions are its predictable side-effect. Checking each candidate in the original equation is the filter that separates real solutions from fake ones. Square, solve, check — all three, every time.