Squaring can invent solutions that were never there. Checking is part of the method, not a formality.
Solve for x
√x + 6 = x
Squaring gives x + 6 = x², so x² − x − 6 = 0 and (x − 3)(x + 2) = 0, offering 3 and −2. Check both in the original: √9 = 3 ✓, but √4 = 2, not −2 ✗. So x = 3 only. B keeps the extraneous root — the one the squaring created. A square root is never negative, so x = −2 could never have worked.
x = 3
one solution
x² = 9
square both sides
x = 3 or x = −3
now two
It turns a statement about one number into a statement about two, so every candidate has to be tested against the original equation. The check is where the invented answer gets caught.
Get the root alone on one side before squaring anything.
√2x − 1 + 4 = 9
the 4 is in the way
√2x − 1 = 5
isolated
2x − 1 = 25
now square
x = 13
check: √25 + 4 = 9 ✓
Squaring too early is the costly error. (√2x − 1 + 4)² is not 2x − 1 + 16 — it has a cross term, and expanding it correctly is far more work than moving the 4 first.
Sometimes the check can be anticipated. In √something = x, the right side must be non-negative, so any negative candidate is doomed before you test it.
√x + 5 = √2x − 1
one each side
x + 5 = 2x − 1
square once — both clear
x = 6
check: √11 = √11 ✓
√x + 7 − √x = 1
isolate one first
√x + 7 = 1 + √x
rearrange
x + 7 = 1 + 2√x + x
square — a radical survives
6 = 2√x, so √x = 3, x = 9
square again; check: 4 − 3 = 1 ✓
The same isolate–square–check routine; the check decides each of these.
No solution at all: √x + 3 + 5 = 2
√x + 3 = −3
isolate
x + 3 = 9, so x = 6
squaring offers a candidate
√9 + 5 = 8, not 2 ✗
no solution
Different: a root alone equal to a negative number can be rejected the moment it is isolated.
Both candidates survive: √4x + 1 = x + 1
4x + 1 = x² + 2x + 1
square both sides
x² − 2x = 0, so x(x − 2) = 0
candidates 0 and 2
x = 0: √1 = 1 ✓ x = 2: √9 = 3 ✓
x = 0 or x = 2
Different: checking does not always throw one away. Keep every candidate that passes.
A negative cube root: 3√2x + 5 = −3
2x + 5 = −27
(−3)³ = −27
2x = −32, so x = −16
check: 3√−27 = −3 ✓
Different: a cube root may equal a negative number, so "no solution" is never automatic here.
3√x − 4 = 2
cube both sides
x − 4 = 8
2³ = 8
x = 12
check: 3√8 = 2 ✓
Cubing does not create extraneous roots, because a cube root accepts negatives and the operation is reversible. Only even roots need the extraneous check — though checking anyway costs nothing.
The usual item is √linear = x or √… + k = c, with "x = a or x = b" offered beside the single correct root to catch anyone who skips the check. The fastest route is isolate, square, solve, then substitute each candidate into the ORIGINAL. The 10-second check: whatever a square root equals must be zero or positive, so reject any candidate that makes that side negative.
Solve for x
√3x + 1 = x − 1
Squaring gives 3x + 1 = x² − 2x + 1, so x² − 5x = 0 and x(x − 5) = 0, offering 0 and 5. Check in the original: at x = 5, √16 = 4 and 5 − 1 = 4 ✓; at x = 0, √1 = 1 but 0 − 1 = −1 ✗. So x = 5 only. A keeps the extraneous root that squaring created.
Hint: the right side must be non-negative, since it equals a square root. That rules out any candidate below 1 before you even substitute.
