Squaring both sides solves it — and invents answers that were never there.
Solve for x
√2x + 3 = x
Squaring gives x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0 and x = 3 or −1. But a square root never returns a negative, so x = −1 fails the check. Only x = 3 survives. B keeps the extraneous root that squaring invented.
Here it already is. When it is not, isolate before squaring.
√2x + 3 = x
√x + 5 = 9
subtract the 5 first
(√x + 5)² is x + 10√x + 25 — the radical is still there, with company. Isolating first avoids the mess entirely.
2x + 3 = x²
the root is gone
x² − 2x − 3 = 0
zero on one side
(x − 3)(x + 1) = 0
factor
x = 3 or x = −1
two candidates
Candidates, not answers. This is the step where the topic gets its reputation.
Squaring turns −1 into 1, so it also turns a false statement into a true one. The squared equation is happy with x = −1; the original never was.
The square root symbol only ever gives a non-negative result.
√2x + 3 = x
the left side is ≥ 0
so x must be ≥ 0
before solving anything
x = −1 was doomed
it fails on sight
Noting that condition at the start saves the check later — though checking anyway costs ten seconds and removes all doubt.
Separate them, then square twice if you must.
√x + 7 = 1 + √x
one on each side
x + 7 = 1 + 2√x + x
square — the right is a binomial
6 = 2√x
the x's cancel
√x = 3, x = 9
square again
(1 + √x)² is 1 + 2√x + x, not 1 + x. Squaring a two-term expression always leaves a middle term.
Check: √16 = 4 and 1 + √9 = 4 ✓
Solve √x + 3 + 7 = 2.
√x + 3 = −5
isolate: − 7
a square root is never negative
no solution — stop here
Different: the answer is decided before any squaring, by the sign of the isolated side.
x + 3 = 25 gives x = 22, and it looks fine — until the check: √25 + 7 = 12, not 2. Squaring turned −5 into 25 and hid the problem.
Solve √5x − 6 = x.
5x − 6 = x² → x² − 5x + 6 = 0
square, zero on one side
(x − 2)(x − 3) = 0 → x = 2 or 3
two candidates
√4 = 2 ✓, √9 = 3 ✓ → x = 2 or x = 3
both pass
Different: checking does not always remove a root — here both candidates are positive and both work.
Solve 3√2x − 3 = −3.
2x − 3 = (−3)³ = −27
cube both sides
2x = −24 → x = −12
3√−27 = −3 ✓
Different: a cube root can be negative, and cubing keeps signs, so it never invents an extraneous root.
Squaring squares everything on that side, the number in front included. Solve 2√x − 1 = x − 4.
4(x − 1) = (x − 4)²
2² = 4, and the right is a binomial
4x − 4 = x² − 8x + 16
keep the middle term −8x
x² − 12x + 20 = 0 → (x − 2)(x − 10) = 0
candidates 2 and 10
x = 10: 2√9 = 6 and 10 − 4 = 6 ✓
keep
x = 2: 2√1 = 2 but 2 − 4 = −2 ✗
extraneous
Leaving the 2 unsquared gives 2x − 2 = x² − 8x + 16, a different quadratic with different roots. Square the coefficient: (2√A)² = 4A.
| Twist | Move |
|---|---|
| x = 3 solves √x + k = x − 1; find k | substitute first: √3 + k = 2, so 3 + k = 4 and k = 1 |
| √x² + 9 = 5 | x² + 9 = 25, x = ±4 — both check, because x² is never negative |
Radical equations are set so that squaring produces one good root and one extraneous root, and the choices include both together. Fastest route: solve, then put each candidate into the original — or just test the choices directly, since substituting into a square root takes seconds.
Squaring never loses a solution — it only ever invents one. So every candidate must face the original.
Solve for x
√x + 6 = x
Squaring gives x² − x − 6 = 0, offering x = 3 or −2. Test both: √3 + 6 = 3 ✓, but √−2 + 6 = 2, not −2. ✗ So x = 3 alone. Squaring can manufacture roots that were never solutions — checking is part of the method, not an extra.
