Equations
Radical Equations and Spurious Solutions
Squaring both sides to remove a radical, why that can create solutions that were never there, and why checking is not optional.
What this page covers
- Solve an equation containing a radical by isolating and squaring
- Explain why squaring can introduce solutions
- Test every candidate in the original equation
- Recognise in advance which candidates are likely to fail
The method and its cost
A radical is removed by raising both sides to the matching power. That works, but for an even power it is not a reversible step, and the irreversibility has a consequence that must be handled.
- Isolate the radical on one side by itself.
- Raise both sides to the power matching the index — square for a square root, cube for a cube root.
- Solve the equation that results.
- Substitute every candidate into the original equation and discard the ones that fail.
Squaring can produce an equation with more solutions than the one you started with. The source notes end the topic with the instruction, in capitals, to check for spurious solutions. Any answer that omits the check is unfinished.
Why squaring adds solutions
Squaring loses sign information. If a = b then certainly a2 = b2; but a2 = b2 is also satisfied when a = -b.
So the squared equation is the union of two problems: the one you asked about and its sign-flipped twin. Its solution set contains yours and may be strictly larger.
A simple illustration: x = 3 has one solution. Squaring gives x2 = 9, which has two, x = 3 and x = -3. The second belongs to x = -3, not to the original.
Odd powers do not have this problem. Cubing is reversible over the reals, because a3 = b3 forces a = b. Equations with cube roots therefore need no check for spurious solutions, though checking is still cheap insurance against arithmetic slips.
The source's two examples
Worked example 1 — √x + 1 = 2x - 3
The radical is already isolated. Square both sides:
The discriminant is 169 - 128 = 41, so
Now the check, which is where the two candidates part company. The right side 2x - 3 must be non-negative, because the left side is a square root.
| Candidate | 2x - 3 | Verdict |
|---|---|---|
| x ≈ 2.425 | ≈ 1.85 > 0 | Valid — and √3.425 ≈ 1.85 agrees |
| x ≈ 0.825 | ≈ -1.35 < 0 | Spurious — a square root cannot be negative |
Only x = 13 + √418 solves the original. The other root solves √x + 1 = -(2x - 3), which is the sign-flipped twin that squaring silently introduced.
Worked example 2 — √x + 2 = x - 4
The source then checks both explicitly.
| Candidate | Left side | Right side | Verdict |
|---|---|---|---|
| x = 7 | √9 = 3 | 7 - 4 = 3 | Valid |
| x = 2 | √4 = 2 | 2 - 4 = -2 | Spurious: 2 ≠ -2 |
The solution is x = 7 alone. The source marks the failing case with a cross and follows it with the instruction to check for spurious solutions.
Predicting failures before checking
A quick sign test filters most spurious candidates without any substitution.
In √A = B, the left side is non-negative by definition of the principal square root. So any solution must satisfy B ≥ 0. Candidates making B negative cannot be solutions.
Both source examples are decided by this test alone. In the first, 2x - 3 < 0 at x ≈ 0.825. In the second, x - 4 < 0 at x = 2. Neither needed the radical to be evaluated.
The radicand must also be non-negative for the expression to be real at all. In √x + 2 that requires x ≥ -2, which both candidates satisfied here, but it will sometimes eliminate candidates on its own.
Two radicals
When two radicals are present, isolate and square twice. Squaring a sum that still contains a radical leaves one behind, which is why the first square must be arranged so only one radical survives.
Worked example — √x + 5 - √x = 1
√9 - √4 = 3 - 2 = 1. Valid.
Isolating √x + 5 before the first squaring was essential. Squaring the original directly gives x + 5 - 2√x(x + 5) + x = 1, which still contains a radical and is worse than what was started with.
Common mistakes
| Mistake | Correct | Why |
|---|---|---|
| Not checking the candidates | Always substitute back | Squaring is not reversible |
| Checking in the squared equation | Check in the original | The squared version accepts spurious roots by construction |
| (2x - 3)2 = 4x2 - 9 | 4x2 - 12x + 9 | The middle term is not optional |
| Squaring before isolating the radical | Isolate first | Otherwise a cross term keeps the radical |
| Discarding a valid root because it looks odd | Test it, do not judge it | 13 + √418 is a fine answer |
| Assuming cube roots need the same check | Odd powers are reversible | Checking is still cheap insurance |
Frequently asked questions
Why does squaring create solutions?
Because squaring destroys sign information. a = b implies a2 = b2, but the reverse does not hold: a2 = b2 also allows a = -b. The squared equation is satisfied by the solutions of both, so it can have more.
Is a spurious solution a mistake?
No. It arises from a legitimate step and its appearance is expected. The mistake is failing to discard it. The check is part of the method, not an optional extra.
Can I tell in advance which will fail?
Often. If the equation is √something = E, the radical is non-negative, so any candidate making E negative must fail. Testing that sign first is a fast filter.
What if there are two radicals?
Isolate one, square, simplify, then isolate the remaining radical and square again. Two squarings mean more scope for spurious solutions, so the check becomes more important, not less.
Source. Handwritten teaching notes, Week 3, pages 8-9.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.