# Radical equation and extraneous solution: -4 or 3 for $\sqrt{12-x}=x$

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?

$$\sqrt{16}$$ 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.

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 $$x-\sqrt{x-2}=8.$$