Radical and Rational Equations, and Extraneous Solutions

Solving is a chain of claims

Solve $2x + 5 = 17$ and write the reason beside each line.

Line	Why it follows from the line above
$2x + 5 = 17$	the equation, assuming some $x$ makes it true
$2x = 12$	equal numbers minus $5$ are still equal
$x = 6$	equal numbers divided by $2$ are still equal

Every line is a claim: if the line above is true for $x$, this line is true for $x$. The chain starts by supposing a solution exists and ends by naming it. That is what solving an equation means.

Each of these steps also runs backwards. From $x = 6$, multiplying by $2$ and adding $5$ lands back on $2x + 5 = 17$. So the chain works in both directions, and $6$ is not only the one candidate — it is certainly a solution.

Steps that only run forwards

Some steps are safe in one direction only.

Step	Forwards	Backwards
add or subtract the same number	✓	✓
multiply or divide by a nonzero number	✓	✓
square both sides	✓	✗
multiply both sides by an expression that could be $0$	✓	✗

Squaring runs forwards. If $a = b$, then $a^2 = b^2$. But $a^2 = b^2$ does not force $a = b$. Both $3$ and $-3$ square to $9$, so $3^2 = (-3)^2$ is true while $3 = -3$ is false.

Multiplying by zero runs forwards. If $a = b$, then $0 \cdot a = 0 \cdot b$. But $0 \cdot 5 = 0 \cdot 7$ is true while $5 = 7$ is false. An expression like $x - 2$ is zero at $x = 2$, so multiplying by it carries the same risk.

Why extraneous solutions appear

When every step runs both ways, the chain proves two things: every solution is on the final list, and everything on the list is a solution.

A one-way step breaks the second half. The chain still proves every solution is on the list — that direction never failed. It no longer proves that every value on the list is a solution.

A value that survives to the end but fails the original equation is an extraneous solution. It was let in at the one-way step, where two different equations became indistinguishable.

$$\sqrt{x + 6} = x \qquad\text{and}\qquad -\sqrt{x + 6} = x$$

Square either one and the result is the same: $x + 6 = x^2$. The squared equation cannot tell them apart, so its solutions include the solutions of both.

So the rule is exact, not a superstition: after a one-way step, substitute every candidate into the original equation. The ones that fail belonged to the other equation.

Radical equations

Isolate the radical, raise both sides to the power that removes it, solve, then check every candidate.

$$\sqrt{x + 6} = x$$

Square both sides.

$$x + 6 = x^2 \quad\Rightarrow\quad x^2 - x - 6 = 0 \quad\Rightarrow\quad (x - 3)(x + 2) = 0$$

The candidates are $3$ and $-2$. Check each in the original.

Candidate	Left side	Right side
$3$	$\sqrt{9} = 3$	$3$	✓
$-2$	$\sqrt{4} = 2$	$-2$	✗

The only solution is $x = 3$. The $-2$ solves $-\sqrt{x + 6} = x$ instead, which is the equation squaring merged with this one.

Cube roots need no such care. Cubing never sends two different numbers to the same result, since $2^3 = 8$ and $(-2)^3 = -8$ differ. So cubing both sides runs both ways and adds nothing.

$$\sqrt[3]{x - 1} = 2 \quad\Rightarrow\quad x - 1 = 8 \quad\Rightarrow\quad x = 9$$

Rational equations

Multiply both sides by the least common denominator to clear the fractions, solve, then throw out any value that makes an original denominator zero.

$$\frac{x^2}{x + 3} = \frac{9}{x + 3}$$

Multiply both sides by $x + 3$, which is zero when $x = -3$.

$$x^2 = 9 \quad\Rightarrow\quad x = 3 \;\text{ or }\; x = -3$$

The value $-3$ makes the original denominators zero, so the original equation does not exist there. The only solution is $x = 3$.

The excluded values are worth writing down before solving. Any candidate on that list is extraneous without further checking.

Worked examples

Common mistakes

Practice problems