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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?

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    $\begingroup$ The function $\sqrt{a}$ is defined to be the non-negative solution to $x^2=a$. $\endgroup$ – robjohn Jan 27 '16 at 16:47
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Note that we define $\sqrt{x^2} = |x|$ so the only real number that satisfies your equation is $9$.

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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.

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  • $\begingroup$ So 4 is a solution to the equation. However it is extraneous. So maybe what I'm wrestling with is what exactly is the definition of Extraneous. Could you say that Extraneous is not Conventional? $\endgroup$ – B Simp Jan 27 '16 at 17:28
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    $\begingroup$ @BSimp extraneous root means a root which may be a solution of the given equation but it does not satisfy some other conditions of the problem like in this case, the equation must have only one root. So we have an extraneous root. You can also check mathsisfun.com/definitions/extraneous-root.html and en.wikipedia.org/wiki/Extraneous_and_missing_solutions $\endgroup$ – SchrodingersCat Jan 27 '16 at 18:38
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    $\begingroup$ @BSimp The number $4$ is a solution to the equation $a = (a - 6)^2$. However, it is not a solution to the original equation $\sqrt{a} = a - 6$ since $\sqrt{4} = 2 \neq -2 = 4 - 6$. $\endgroup$ – N. F. Taussig Jan 27 '16 at 22:10
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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.

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