Extraneous Solution
In short: An extraneous solution is a number your algebra hands you that then fails inside the starting equation, typically a leftover from squaring or from clearing a variable denominator.
Algebra 2AlgebraA-REI.A.2
What is Extraneous Solution?
An extraneous solution is an impostor: it appears at the end of correct-looking steps, yet it refuses to work in the problem you began with.
The trouble comes from steps that are not reversible. Squaring both sides can turn a false statement such as \(3 = -3\) into the true statement \(9 = 9\), and multiplying by an expression containing \(x\) can bring in a value that makes a denominator zero. Substituting every candidate into the original equation is the only sure filter.
How to spot it
- The equation has a square root with the variable inside, or a variable in a denominator.
- You squared both sides, or multiplied through by an expression that contains x, to clear it.
- Solving produced two candidate values and the problem warns you to check or to find all valid answers.
- A candidate would make a denominator zero or a square root equal to a negative number.
The method
- Solve the equation with the usual steps and list every candidate value.
- Plug each candidate into the original equation, not into a later line of your work.
- Discard any candidate that zeroes a denominator, forces a negative under a root, or leaves the two sides unequal.
- State only the surviving values as solutions; if none survive, write no solution.
Extraneous Solution: solved example
Problem
Solve \(\sqrt{2x + 9} = x + 3\) and identify any extraneous solution.
Solution
Squaring turns the equation into \(2x + 9 = x^2 + 6x + 9\), which reduces to \(x(x + 4) = 0\). That leaves two suspects: \(x = 0\) and \(x = -4\).
Test \(x = 0\): the left side is \(\sqrt{9} = 3\) and the right side is \(3\), a match. Test \(x = -4\): the left side is \(\sqrt{1} = 1\) while the right side is \(-1\), a mismatch. Hence \(x = 0\) is valid and \(x = -4\) is extraneous.
Try it
Question 1
Solve \(\sqrt{x + 6} = x\).
Show the answer
Squaring gives \(x^2 – x – 6 = 0\), so \(x = 3\) or \(x = -2\). Since \(\sqrt{4} = 2 \neq -2\), the value \(-2\) is extraneous; the answer is \(x = 3\).
Question 2
Solve \(\frac{x}{x – 5} = \frac{5}{x – 5} + 2\).
Show the answer
Multiply by \(x – 5\): \(x = 5 + 2(x – 5)\), so \(x = 5\). But 5 zeroes both denominators, so it is rejected and the equation is unsolvable.
Extraneous Solution vs. Rational Expression
An extraneous value is not the same as an excluded value. Excluded values belong to a fraction and are known before you start, because they zero its denominator. A false answer is discovered afterward, through the check, and it can arise from radicals as well as from fractions.
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