Key takeaway: An extraneous solution is a value that appears while solving an equation but fails when substituted into the original equation, usually after squaring or clearing denominators.
Grade 11 – Algebra 2A-REI.A.2Expressions and Equations
Definition of Extraneous Solution
An extraneous solution is a false answer that the solving process hands you. Every line of algebra can be correct, yet the number at the end still fails in the equation you began with.
The trouble starts with steps that cannot be undone. Squaring turns the false claim \(-4=4\) into the true claim \(16=16\), so a wrong value sneaks through. Multiplying both sides by an expression that might be zero can do the same in fractional equations. For that reason, the last move is always a substitution into the starting equation.
Formula and notation
No special symbol exists. Write the check as a substitution into the starting equation and compare the two sides:
\[\text{left side}(x_0) \neq \text{right side}(x_0) \implies x_0 \text{ is rejected.}\]
Cross out the rejected value and list only the numbers that survive.
Key facts
- Radical equations and equations with variable denominators are the usual sources of false answers.
- A value is rejected when it makes a denominator zero or sets a principal root equal to a negative.
- Checking in the original equation is required, not optional, for these equation types.
- An equation can end up with no valid answers even though the algebra produced candidates.
Extraneous Solution example
Solve \(\sqrt{x+10}=x-2\) and decide which candidates are false.
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Square both sides: \(x+10=x^2-4x+4\), which rearranges to \(x^2-5x-6=0\).
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Factor into \((x-6)(x+1)=0\), so the candidates are 6 and \(-1\).
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Test 6: the left side is \(\sqrt{16}=4\) and the right side is \(6-2=4\). They agree.
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Test \(-1\): the left side is \(\sqrt{9}=3\) but the right side is \(-1-2=-3\). They disagree.
Only \(x=6\) solves the equation; \(x=-1\) is an extraneous solution.
Common mistake
Watch out
Lina solves \(\frac{x}{x-5}=\frac{5}{x-5}+2\), gets \(x=5\) after clearing denominators, and reports 5 as the answer without substituting.
Why it is wrong
At \(x=5\) both denominators are zero, so the starting equation is undefined there. The candidate is an extraneous solution, and the equation has no valid answer. Always test candidates in the starting equation.
Related terms and lessons
- Algebra 2 · Rational Expressions and EquationsLessonWorksheetQuiz
- Algebra 2 · Radicals and Rational ExponentsLessonWorksheetQuiz
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