Squaring both sides is a useful way to remove a square root, but it can introduce answers that do not belong to the original equation. The fix is simple: substitute every candidate into the original equation and reject any that fail.
This skill belongs to algebraic equations and inequalities. It builds on the same “test the answer” habit used in solving an equation involving an absolute value.
Where does the false answer come from?
A square root symbol √ means the non-negative root only. So √(x + 6) is never negative, and an equation √(x + 6) = x can only be true when x itself is zero or positive.
Squaring gives x + 6 = x², and this new equation is also satisfied by any x that would make √(x + 6) equal to −x. Those values satisfy the squared equation but not the original.
Method, step by step
- Isolate the surd on one side.
- Square both sides, expanding any bracket carefully.
- Rearrange into a quadratic and solve it.
- Substitute each root into the original equation.
- Reject any root that makes the two sides unequal, and state your final answer.
Worked example
Solve √(x + 6) = x.
Step 1, surd is already isolated.
Step 2, square both sides: x + 6 = x².
Step 3, rearrange and solve: x² − x − 6 = 0, so (x − 3)(x + 2) = 0. The candidates are x = 3 and x = −2.
Step 4, check x = 3: √(3 + 6) = √9 = 3. The right side is 3. It works.
Step 4, check x = −2: √(−2 + 6) = √4 = 2. The right side is −2. Since 2 ≠ −2, this fails.
Answer: x = 3 only. The root x = −2 is extraneous.
Quick reasoning check: the right side x must be non-negative because it equals a square root. That alone rules out −2 before any substitution.
The mistake to watch for
The error is to stop at the quadratic and report both solutions.
Mistaken answer: x = 3 or x = −2
The student solved the quadratic correctly but never returned to the original equation.
The correction is to treat the substitution check as the final step of the method, not an optional extra. A second error is the opposite one: rejecting a root without testing it. If you reject, show the value that fails, as in √4 = 2 but the right side is −2.
Check yourself
Try these, then open each answer.
1. Solve √(3x + 4) = x.
Show answer
Square: 3x + 4 = x², so x² − 3x − 4 = 0, which gives (x − 4)(x + 1) = 0. Candidates: 4 and −1.
Check 4: √16 = 4, matches. Check −1: √1 = 1, but the right side is −1, so it fails.
x = 4
2. Solve √(5x − 1) = x + 1.
Show answer
Square: 5x − 1 = x² + 2x + 1, so x² − 3x + 2 = 0, which gives (x − 1)(x − 2) = 0.
Check 1: √4 = 2 and 1 + 1 = 2, matches. Check 2: √9 = 3 and 2 + 1 = 3, matches.
x = 1 or x = 2 (both valid)
3. Solve √(x + 7) − 1 = x.
Show answer
Isolate the surd: √(x + 7) = x + 1. Square: x + 7 = x² + 2x + 1, so x² + x − 6 = 0, which gives (x + 3)(x − 2) = 0.
Check 2: √9 − 1 = 2, matches. Check −3: √4 − 1 = 1, but the right side is −3, so it fails.
x = 2
Where this leads next
The same idea of restricting what an answer is allowed to be continues in stating restrictions before manipulating fractions. Try the mixed practice set when you are ready, and use the non-calculator working trainer for clean exact substitutions.
Students who solve the algebra fluently often lose marks only at the checking stage. A teacher in online one-to-one Additional Mathematics tuition can help you build a checking routine that fits your own working style.