I have a radical equation to solve:


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?



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


$12-x=x^2$ # 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.

| cite | improve this answer | |

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.