8.1 Extraneous Roots
This chapter is about a single hazard: some of the steps used to solve an equation can create solutions that do not solve the original. Such a value is called an extraneous root, and the only defence is to substitute every answer back into the equation you started with.
Two steps are the usual culprits. Clearing denominators can produce a value that makes an original denominator zero, where the equation was never defined. Squaring both sides destroys sign information, because \(u=v\) and \(u=-v\) both lead to \(u^{2}=v^{2}\) — so the squared equation cannot tell them apart and answers to the wrong question as well as the right one.
Neither step loses solutions; both may add them. So every solution of the original is among the values found, and the check is simply to discard the impostors.
\begin {align*} \text {So}\quad \frac {(x-2)+(x-3)}{(x-2)(x-3)} & = \frac {3x-8}{(x-2)(x-3)}\\\\ \implies \quad \frac {2x-5}{(x-2)(x-3)} & = \frac {3x-8}{(x-2)(x-3)}\\\\ \implies \quad 2x-5 & = 3x-8\\\\ \implies \quad x & = 3\\ \end {align*}
Now check. Substituting \(x=3\) into the original equation gives \(\frac {1}{0}\), which is undefined — \(x=3\) was never in the domain of either side. It arrived only when the denominators were cleared, so it is extraneous and must be discarded.
Since it was the only candidate, the equation has no solution.
Note 8.2. Note what the working actually established: if a solution exists then it must be \(3\). That is a genuine deduction, and it remains true. The check then shows no such solution exists. An implication chain of this kind gives candidates, not answers, and the last line of the working is never the last line of the problem.
- (a).
- \(\sqrt {x}=x-2\)
- (b).
- \(\sqrt {2-x}=-3\)
- (c).
- \(2x+7=3\sqrt {3-2x}\)
- (d).
- \(\sqrt {5x-1}=\sqrt {x}+1\)
Solution. (a). \(\sqrt {x}=x-2\). Squaring, \[x=(x-2)^{2}=x^{2}-4x+4 \implies x^{2}-5x+4=0 \implies (x-1)(x-4)=0,\] so \(x=1\) or \(x=4\).
Checking: at \(x=1\), the left side is \(\sqrt {1}=1\) and the right is \(1-2=-1\). These differ, so \(x=1\) is extraneous — it solves \(\sqrt {x}=-(x-2)\), the other equation that squaring cannot distinguish. At \(x=4\), \(\sqrt {4}=2\) and \(4-2=2\), so \(x=4\) is a genuine solution.
\[\boxed {x=4}\]
(b). \(\sqrt {2-x}=-3\). No work is required. The square-root symbol denotes the non-negative root, so the left side is never negative and can never equal \(-3\). There is no solution.
Squaring here would give \(2-x=9\), hence \(x=-7\), which is entirely extraneous: \(\sqrt {2-(-7)}=\sqrt {9}=3\), not \(-3\). This is the clearest possible illustration of why the check is not optional.
(c). \(2x+7=3\sqrt {3-2x}\). Squaring, \begin {align*} (2x+7)^{2}&=9(3-2x)\\ 4x^{2}+28x+49&=27-18x\\ 4x^{2}+46x+22&=0\\ 2x^{2}+23x+11&=0\\ (2x+1)(x+11)&=0, \end {align*}
so \(x=-\frac {1}{2}\) or \(x=-11\).
Checking: at \(x=-\frac {1}{2}\), the left side is \(-1+7=6\) and the right is \(3\sqrt {3+1}=3(2)=6\). Genuine. At \(x=-11\), the left side is \(-22+7=-15\) while the right is \(3\sqrt {3+22}=3(5)=15\). The two are negatives of each other — the signature of an extraneous root from squaring — so it is discarded.
\[\boxed {x=-\frac {1}{2}}\]
(d). \(\sqrt {5x-1}=\sqrt {x}+1\). One square root sits on each side, so squaring once will not clear them both. Squaring the first time, \begin {align*} 5x-1 & = \left (\sqrt {x}+1\right )^{2}\\ & = x+2\sqrt {x}+1, \end {align*}
which rearranges to \[4x-2=2\sqrt {x} \qquad \text {that is}\qquad 2x-1=\sqrt {x} .\] Squaring a second time, \begin {align*} (2x-1)^{2} & = x\\ 4x^{2}-4x+1 & = x\\ 4x^{2}-5x+1 & = 0\\ (4x-1)(x-1) & = 0, \end {align*}
so \(x=\frac {1}{4}\) or \(x=1\).
Checking: at \(x=\frac {1}{4}\) the left side is \(\sqrt {\frac {5}{4}-1}=\sqrt {\frac {1}{4}}=\frac {1}{2}\), while the right side is \(\frac {1}{2}+1=\frac {3}{2}\). These differ, so \(x=\frac {1}{4}\) is extraneous. At \(x=1\) the left side is \(\sqrt {4}=2\) and the right is \(1+1=2\), so \(x=1\) is genuine.
\[\boxed {x=1}\]
The rejection could have been seen earlier. The intermediate equation \(2x-1=\sqrt {x}\) has a non-negative right side, so it requires \(2x-1\geq 0\), that is \(x\geq \frac {1}{2}\) — and \(\frac {1}{4}\) already fails it. Recording the condition at the moment of squaring often disposes of the impostor before any checking is needed.
Note 8.4. A pattern runs through all four. Whenever a value is rejected, the left and right sides turn out to be equal in size and opposite in sign — \(1\) against \(-1\) in (a), \(-15\) against \(15\) in (c), \(\frac {1}{2}\) against \(\frac {3}{2}\) in (d) being the one exception, where squaring twice muddied it further.
That is exactly what squaring does: it merges \(u=v\) with \(u=-v\). The roots of the squared equation are the solutions of both, and the check sorts them.
A useful precaution: before squaring, note any restriction the original equation already imposes. In (a), \(\sqrt {x}=x-2\) needs the right side to be non-negative, so \(x\geq 2\) — which rules out \(x=1\) before any checking is done. In (c), \(2x+7=3\sqrt {3-2x}\) needs \(2x+7\geq 0\), so \(x\geq -\frac {7}{2}\), which rules out \(x=-11\) immediately. Recording that restriction first often saves the work at the end.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.