How do you solve $\sqrt{12 - n} = n$ $sqrt(12-n)=n$ and find any extraneous solutions?

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Answer 1 · 2016-08-28T20:46:02

$n = 3$ $n=3$

Given:

$\sqrt{12 - n} = n$ $sqrt(12-n) = n$

Square both sides (noting that this may introduce spurious solutions):

$12 - n = n^{2}$ $12-n = n^2$

This derived equation may (and actually does) have solutions which are not solutions of the original equation.

Subtract $(12 - n)$ $(12-n)$ from both sides to get:

$0 = n^{2} + n - 12 = (n + 4) (n - 3)$ $0 = n^2+n-12 = (n+4)(n-3)$

Hence $n = - 4$ $n=-4$ or $n = 3$ $n=3$

$n = - 4$ $n=-4$ is not a solution of the original equation since:

$\sqrt{12 - (- 4)} = \sqrt{16} = 4 \neq - 4$ $sqrt(12-(-4)) = sqrt(16) = 4 != -4$

$n = 3$ $n=3$ is a valid solution of the original equation:

$\sqrt{12 - 3} = \sqrt{9} = 3$ $sqrt(12-3) = sqrt(9) = 3$

How do you solve sqrt(12-n)=n√12−n=n sqrt(12-n)=n and find any extraneous solutions?