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I have a radical equation to solve:

$\sqrt{12-x}=x$

Before checking for extraneous solutions I arrived at -4 and 3. My textbook says the only solution is 3 but surely it's -4 too?

$\sqrt{12-(-4)}=-4$

$\sqrt{16}=-4$

This is correct no? A square root of 16 is -4?

$\sqrt{12-x}=x$

$12-x=x^2$ # remove radical by squaring both sides

$-x^2-x+12=0$ # move everything to one side

$x^2+x-12=0$ # multiple both sides by -1 to get a positive exponential term

$x^2+4x-3x-12=0$ # split into groups (what's the conventional name of this step?)

$x(x+4)-3(x+4)=0$ # not sure the name of this step? "pre" factoring?

$(x+4)(x-3)$ # factor into groups

$x+4=0$; $x=-4$

$x-3=0$; $x=3$

Why is it that 3 is the only solution?

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$\sqrt{16}$ is usually notation for the $\textit{positive}$ solution of $x^2=16$. So

$$y^2=x\impliedby y=\sqrt{x}$$

but the inverse implication is generally not true. In general what you have is $$y^2=x\iff y=\pm\sqrt{x}$$

So $\sqrt{12-x}$ is positive by definition.

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Always remember that $\sqrt{x^2}=|x|$ ant it is not $\pm x$. Roots of $x^2=9$ are $\pm 3$ but $\sqrt{9}$ is not $\pm 3$, it is only 3. $\sqrt{7}= +\sqrt{7},$ and not $-\sqrt{7}$.

Whenever you square an equation on both sides, the two sides must be posotive. If not declare them so, and finally take it into account.

If real ($12> x$), $\sqrt{12-x}$ is always positive. So declare $x>0$ and square the the given equation, you get $x^2+x-12=0$. Finally out of two roots $x=-4, 3$ only 3 is the correct root and the other one is extraneous.

An interesting example is $x+\sqrt{x-2}=8$, here only $x=6$ is a root. Actually, $x=11$ is the root of $x-\sqrt{x-2}=8.$

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