What is the extraneous solution of $$\sqrt a=a-6$$ The roots are $9$ and $4$. So I'm assuming that $4$ is the extraneous solution because when you plug it in to the equation you wind up with $2=-2$. However, isn't the square root of $4$ both $-2$ & $2$? What am I missing here?
$\endgroup$ 13 Answers
$\begingroup$Note that we define $\sqrt{x^2} = |x|$ so the only real number that satisfies your equation is $9$.
$\endgroup$ $\begingroup$Your statement is correct that both $2$ and $-2$ when squared yield $4$.
However in general, by the square root function, we indicate the positive square root i.e. $\sqrt{x^2}=|x|$.
So you know that $9$ is the only real solution of your equation and $4$ is the extraneous solution.
$\endgroup$ 3 $\begingroup$By convention, the square root function $\sqrt{x}$ denotes the positive square root, so $a - 6 \geq 0$, implying that $a = 9$ is the correct solution, while $a = 4$ is the extraneous solution.
$\endgroup$
