MISSION · Algebra
Solutions That Aren't Solutions
Why squaring both sides can hand you an answer that doesn't work
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The idea in one sentence
Some steps in solving an equation are not reversible, so they can produce values that satisfy the rearranged equation but not the one you started with — and the only way to know is to check.
A worked case
\[\sqrt{x + 6} = x\]Square both sides:
\[x + 6 = x^2 \Rightarrow x^2 - x - 6 = 0 \Rightarrow (x-3)(x+2) = 0\]So \(x = 3\) or \(x = -2\). Check them in the original:
- \(x = 3\): \(\sqrt{9} = 3\). True.
- \(x = -2\): \(\sqrt{4} = 2\), not \(-2\). False.
\(x = -2\) is an extraneous solution. It isn’t an arithmetic mistake — the algebra was correct at every step. It’s a solution to \(x + 6 = x^2\), which is a slightly different equation from the one asked.
Why squaring does this
Squaring destroys information about sign. \(3\) and \(-3\) square to the same thing, so the squared equation can no longer tell them apart, and it says yes to both. The square root symbol means the non-negative root, so the original equation was quietly demanding \(x \geq 0\) all along — a condition that vanished the moment you squared.
The general rule: \(a = b\) implies \(a^2 = b^2\), but not the other way round. Any step that only runs one way can add solutions.
The other main source: multiplying by an expression
\[\frac{1}{x-2} + 3 = \frac{x}{x-2}\]Multiply through by \(x - 2\):
\[1 + 3(x-2) = x \Rightarrow 3x - 5 = x \Rightarrow x = 2.5\]That one checks out. But watch what happens with
\[\frac{x}{x-2} = \frac{2}{x-2}\]Multiplying gives \(x = 2\) — which makes both denominators zero, so it was never allowed. The equation has no solution at all.
Here the giveaway comes earlier than the check: \(x = 2\) is excluded from the domain before any solving begins. Note the excluded values first, and any answer that hits one gets crossed off on sight.
Which operations are safe
Always safe: adding or subtracting the same thing from both sides; multiplying or dividing by a nonzero constant.
Can add solutions: squaring (or any even power); multiplying by an expression containing the variable.
Can lose solutions: dividing by an expression containing the variable. From \(x^2 = 5x\), cancelling an \(x\) gives \(x = 5\) and silently throws away \(x = 0\). Factor instead of cancelling — \(x(x-5) = 0\) keeps both.
Also worth checking: logarithms, where \(\log(x) + \log(x-3) = 1\) can produce a negative \(x\) that no logarithm will accept.
The habit
If your solving involved squaring, multiplying by a variable expression, or logs, substituting back is part of the method, not optional diligence. It takes ten seconds and it’s where marks are routinely lost.
Common mix-ups
- Checking in the squared equation instead of the original, which of course accepts both.
- Rejecting a negative answer on principle. \(\sqrt{x+6} = -x\) has \(x = -2\) as a perfectly good solution. Check, don’t guess.
- Cancelling a variable factor and losing a root.
- Skipping the check when both answers “look fine”. Looking fine is not the test.
Check yourself
Solve √(x + 6) = x and say which root must be rejected. Then: in 1/(x−2) + 3 = x/(x−2), which value is excluded before you start?
Squaring gives x = 3 and x = −2; only x = 3 survives, since √4 is 2 and not −2. In the second, x = 2 makes the denominator zero, so it’s excluded from the outset.
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