Radical Equations Checker

Square both sides, solve the resulting polynomial, then verify each candidate in the original equation.

Worked example — Radical Equations Checker

Original Equation	1x+1
Squared Equation	(1x+1)²
Quadratic Form	0

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Radical Equations Checker

An extraneous root is a solution that appears when you square both sides of an equation and does not solve the original. It is not an arithmetic slip: squaring is not reversible, so the squared equation is genuinely a different equation, satisfied by more values than the one you started with.

What each symbol means

a, b: the numbers under the radical, giving √(ax + b) — 2 and 3 at the defaults.
c, d: the line on the right, cx + d — both 1 here, so the equation is √(2x+3) = x+1.
candidate: a root of the squared equation. There are two here, x ≈ −1.4142 and x ≈ 1.4142.
valid: a candidate that survives substitution into the original equation. Only x ≈ 1.4142 does.

Where the formula comes from

The radical √(2x+3) is by definition the non-negative square root. So the original equation quietly carries a condition the notation does not show: the right-hand side must not be negative.
Squaring both sides throws that condition away. √(2x+3) = x+1 and 2x+3 = (x+1)² are not equivalent — the second is also satisfied whenever √(2x+3) = −(x+1), the negative branch.
So the squared equation’s roots are the solutions of both branches mixed together. Substituting each candidate back is not a formality, it is the step that separates them: at x ≈ −1.4142 the right side is negative, so it solves the negative branch and not the original.

Assumes: Real numbers only, a square root meaning the principal (non-negative) one, and a right-hand side that is linear. Allow complex numbers and the whole notion of an extraneous root changes shape.
Breaks when: Squaring can also lose nothing at all, which makes the habit easy to forget: if the right-hand side happens to be non-negative wherever the radical is defined, every candidate is valid and checking looks like wasted effort. The check is insurance, and the premium is only visible on the occasions it pays out.

Some radical equations can be rejected before you start 🖖

The principal square root is never negative — that is part of its definition, not a convention you can bend. So √(x − 5) = −2 has no solution at all, and you can say so without doing any algebra: the left side cannot reach a negative value for any x whatsoever. Square both sides anyway and you get x = 9, a confident, wrong answer with nothing to warn you. The same pre-check applies to the radicand: it must be at least 0, so any candidate that makes it negative was never in the running. Checking the shape of an equation before solving costs one glance and catches errors that no amount of careful algebra will.

Why you must check every answer 🖖

Squaring both sides is a one-way street: because both 3 and −3 square to 9, that step can admit answers the original equation never allowed. The root symbol returns only the non-negative value, so the right side cx+d must be ≥ 0. Takeaway: substitute every candidate back into the untouched original equation — this tool does it automatically and flags the impostors.

Extraneous roots solve the mirror equation 🖖

A rejected root is not meaningless — it is the exact solution of the sign-flipped twin √(ax+b) = −(cx+d). Squaring erases the minus sign, so both equations share one polynomial, and its candidates split cleanly: those with cx+d ≥ 0 belong to the original, the rest to its mirror. That is why every extraneous point lands precisely on the dashed negative branch y = −√(ax+b) in the plot.

Practice

Check yourself

Predict the answer first, then use the controls above to find out. Reveal only after you have committed to a guess — that is what makes it practice.

All three presets report exactly one extraneous root. Can you make the tool report none at all?

Show answer

Easily. Enter a = 1, b = 0, c = 0.5, d = 0 — that is √x = x/2 — and the panel lists candidates x ≈ 0 and x ≈ 4, puts both under Valid Roots, and prints “None” for extraneous. √x = x does the same with 0 and 1. Squaring only ever *risks* adding a solution; it adds one exactly when a candidate makes the right-hand side negative, because such a point solves y = −√(ax+b), the dashed branch the tool plots, rather than the equation you started from. No candidate goes negative here, so nothing is lost.
The default preset squares to 1x² + 0x − 2 = 0 and offers the symmetric pair ±1.4142, of which the tool rejects −1.4142. Now enter a = 6, b = 12, c = 1, d = 3. You get the same shape of quadratic — 1x² + 0x − 3 = 0, candidates ±1.7321 — but this time the tool accepts both. What decides it?

Show answer

The sign of the right-hand side at each candidate, and nothing else. In the default the line is x + 1, which at x = −1.4142 comes to −0.4142; a square root cannot equal a negative number, so that point lies on the dashed branch. In the second the line is x + 3, which at x = −1.7321 is still +1.2679, and √(6(−1.7321) + 12) = √1.6077 = 1.2679 — a genuine solution, negative x and all. The linear term vanishes in both cases because a = 2cd (2 = 2·1·1, and 6 = 2·1·3), which is a fact about the algebra and says nothing whatever about which roots survive. Extraneous roots are not the ones that look wrong, and not the ones outside the domain — they are the ones where the line has dropped below the axis.

Problem solved in full

The extraneous root case for √(x − 1) = x − 2 7 steps
Solve √(x − 1) = x − 2, and name the step that manufactured the root which does not solve it. This is the extraneous root case: a = 1, b = −1, c = 1, d = −2.
1. The symbol on the left is the principal square root, so its value is never negative. That constrains the right-hand side before any algebra happens: only x from 2 upwards can possibly work.
  
  $x - 1 = x - 2, \cdot \geq 0 ⟹ x \geq 2$
2. Squaring is where the information leaves. Equal quantities have equal squares, but equal squares only pin the quantities down to within a sign. Everything after this line answers a wider question than the one you were asked.
  
  $u = v ⟹ u^{2} = v^{2}, u^{2} = v^{2} ⟺ u = \pm v$
3. Expanding the right-hand side leaves a quadratic, and that is the structural reason a second root has room to appear — the original equation had one branch, the squared one has two.
  
  $x - 1 = (x - 2)^{2} = x^{2} - 4 x + 4 ⟹ x^{2} - 5 x + 5 = 0$
4. The discriminant is 25 − 20 = 5, so both candidates are irrational. These two numbers are exactly what the panel lists under polynomial candidates, and both are genuine roots of the squared equation.
  
  $x = \frac{5 \pm 25 - 20}{2} = \frac{5 \pm 5}{2} \approx 1.382, 3.618$
5. Test the larger candidate. The radicand is 2.618 and its principal root is 1.618; the right-hand side is also 1.618. Both sides agree, so 3.618 survives.
  
  $3.618 - 1 = 2.618 = 1.618 = 3.618 - 2$
6. Now the smaller one. The left side comes out at 0.618 and the right side at −0.618: same size, opposite sign. That is not a rounding miss — 1.382 is an exact root of √(x − 1) = −(x − 2), the lower half of the sideways parabola that the radical sign discards, which is why it lands in the extraneous row.
  
  $1.382 - 1 = 0.382 = 0.618, 1.382 - 2 = - 0.618$
7. Notice what the domain check you were probably taught would have said here. At any root of the squared equation the radicand equals (x − 2)², a square, so it is non-negative at both candidates and rejects neither. On this whole family of equations the radicand test does no work at all; only the sign of the right-hand side does.
  
  $x^{2} - 5 x + 5 = 0 ⟺ x - 1 = (x - 2)^{2} \geq 0$
Answer

x ≈ 3.618, and 1.382 is not a near miss — it is an exact solution of a different equation. Ask when the trouble can be avoided altogether. Keep the radicand at x − 1 and let the right-hand side be x + d, so squaring gives f(x) = x² + (2d − 1)x + (d² + 1) = 0 with discriminant −4d − 3: above d = −0.75 there are no real candidates at all. A candidate survives the sign test when x ≥ −d, the parabola's vertex always sits to the right of −d, so both roots clear the cut-off exactly when f(−d) ≥ 0 — and f(−d) collapses to d + 1. The family is therefore honest only on the narrow window −1 ≤ d ≤ −0.75: raise d past −0.75 and the roots vanish, drop it below −1 and one of them turns extraneous. At d = −2 you are far outside that window, which is why exactly one candidate had to fail.

$Δ = - 4 d - 3 \geq 0, f (- d) = d + 1 \geq 0 ⟹ - 1 \leq d \leq - \frac{3}{4}$

References (2)

Why squaring both sides produces solutions the original equation does not have: J. S. Zelkowski, "Making Sense of Extraneous Solutions." The Mathematics Teacher 106(6), 452–458, 2013.
The idea has a longer history than the textbook treatment suggests: K. R. Manning, "Historically Speaking — A History of Extraneous Solutions." The Mathematics Teacher 63(2), 165–175, 1970.

By the 42Math team · Updated 2026-05-21

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

single-root case - Single valid solution case for sqrt(x+4)=x.
extraneous root case - Squaring introduces a candidate that fails original-domain check.
two-candidate case - Two algebraic candidates appear before radical-domain filtering.