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Solving Radical Equations

A radical equation has the variable trapped inside a square root, like x+5=4\sqrt{x + 5} = 4 or 2x−3=x−1\sqrt{2x - 3} = x - 1. The strategy is short: get the radical alone on one side, then square both sides — squaring is the opposite of a square root, so the radical disappears and you're left with an ordinary equation.

But radical equations come with a catch no other equation type has: squaring both sides can manufacture answers that don't actually work. These fakes are called extraneous solutions, and checking every candidate in the original equation isn't optional — it's part of the solve.

The three-step method

Step 1: isolate the radical. Move everything else to the other side first, so the square root sits alone. If you square while a +7+ 7 is still attached, you'd have to expand (x+7)2(\sqrt{x} + 7)^2 with FOIL — messy and error-prone. Isolate first and squaring is clean.

Step 2: square both sides. The left side loses its radical: (x+5)2=x+5(\sqrt{x + 5})^2 = x + 5. Square the entire right side too — if the right side is x−1x - 1, you get (x−1)2=x2−2x+1(x - 1)^2 = x^2 - 2x + 1, not x2+1x^2 + 1.

Step 3: solve what's left and check every answer in the original equation. If squaring produced a quadratic, expect up to two candidates — and be ready to throw one out.

Why extraneous solutions happen

Squaring erases sign information. The false statement −3=3-3 = 3 becomes the true statement 9=99 = 9 after squaring — so a squared equation can be true even when the original wasn't. Any candidate that only satisfies the squared version is extraneous.

In practice, extraneous solutions show up when the candidate would force a square root to equal a negative number. A square root symbol always means the nonnegative root: x+2=x\sqrt{x + 2} = x can't have a negative xx as a solution, because the left side is never negative. That's exactly the check to run: plug each candidate into the original equation and see if both sides truly match.

Spotting a no-solution equation early

If, after isolating, the radical is set equal to a negative number — like x−1=−3\sqrt{x - 1} = -3 — stop. A square root can't equal a negative, so there is no solution. If you square anyway, the algebra will happily hand you a candidate, and the check will reject it. Recognizing the situation before squaring saves the whole trip.

Worked examples

Example 1: a basic radical equation

Solve 2x−3=5\sqrt{2x - 3} = 5.

The radical is already isolated, so square both sides2x−3=252x - 3 = 25
Add 33 to both sides2x=282x = 28
Divide both sides by 22x=14x = 14
Check in the original: 2(14)−3=25=5\sqrt{2(14) - 3} = \sqrt{25} = 5 — it works

Answer: x=14x = 14

Example 2: isolate the radical first

Solve x−1+7=10\sqrt{x - 1} + 7 = 10.

Subtract 77 from both sides to isolate the radicalx−1=3\sqrt{x - 1} = 3
Square both sidesx−1=9x - 1 = 9
Add 11 to both sidesx=10x = 10
Check: 10−1+7=3+7=10\sqrt{10 - 1} + 7 = 3 + 7 = 10 — it works

Answer: x=10x = 10

Example 3: an extraneous solution appears

Solve x+6=x\sqrt{x + 6} = x.

Square both sidesx+6=x2x + 6 = x^2
Move everything to one sidex2−x−6=0x^2 - x - 6 = 0
Factor(x−3)(x+2)=0(x - 3)(x + 2) = 0
Two candidatesx=3orx=−2x = 3 \quad \text{or} \quad x = -2
Check x=3x = 3: 9=3\sqrt{9} = 3 — it works
Check x=−2x = -2: 4=2\sqrt{4} = 2, but the right side is −2-2. Since 2≠−22 \neq -2, this candidate is extraneous — discard it

Answer: x=3x = 3 only (x=−2x = -2 is extraneous)

Example 4: no solution at all

Solve x−1=−3\sqrt{x - 1} = -3.

A square root is never negative, so no xx can make the left side equal −3-3
If you square anyway, you get a candidatex−1=9  ⇒  x=10x - 1 = 9 \;\Rightarrow\; x = 10
Check x=10x = 10: 9=3\sqrt{9} = 3, not −3-3 — the candidate fails

Answer: No solution

Try one yourself

Common questions

Do I really have to check my answers every time?

Yes — for radical equations the check is part of the method, not an afterthought. Squaring both sides can create extraneous solutions, and the only way to catch them is substituting each candidate back into the original equation. Graders know this and love putting an extraneous solution among the answer choices.

What if the equation has a cube root instead of a square root?

Cube both sides instead of squaring. Bonus: cube roots can equal negative numbers, and cubing doesn't erase sign information, so cube-root equations don't produce extraneous solutions. The check is still a good habit, but nothing should fail it.

What if there are two radicals in the same equation?

Isolate one radical and square both sides — that clears the first one. Usually a radical remains; isolate it and square again. Two rounds of squaring means extra chances for extraneous solutions, so checking at the end matters even more.

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