# Math Help - Solve equation

2. ## Re: Solve equation

Originally Posted by Mhmh96
Question in Latex: $x = 12-\sqrt{12-\sqrt{x}}$

What have you tried? A good start would be to rearrange the equation to isolate the radical term so it's easier to square both sides

From the domain we know that $x \geq 0 \text{ and } \sqrt{x} \leq 12 \Leftrightarrow x \leq 144$. Overall this means $0 \leq x \leq 144$

Edit: Don't forget to test potential solutions in the original equation to check if they are spurious or not

3. ## Re: Solve equation

I tried to square both side but still not able to
determine to the right answer which is 9

4. ## Re: Solve equation

Hello, Mhmh96!

$\text{Solve for }x\!:\;\;x \;=\;12 - \sqrt{12-\sqrt{x}}$

Let $u \,=\, \sqrt{x}\quad\Rightarrow\quad x \,=\,u^2$

We have: . $u^2 \:=\:12 - \sqrt{12-u} \quad\Rightarrow\quad \sqrt{12-u} \;=\;12 - u^2$

Square both sides: . $\left(\sqrt{12-u}\right)^2 \;=\; \left(12-u^2\right)^2$

. . $12-u \;=\;144 - 24u^2 + u^4 \quad\Rightarrow\quad u^4 - 24u^2 + u + 132 \:=\:0$

which factors: . $(u-3)(u+4)(u^2-u-11) \:=\:0$

We have: . $\begin{Bmatrix} u-3 \:=\:0 & \Rightarrow & u \:=\:3 \\ \\[-3mm] u+4\:=\:0 & \Rightarrow & u \:=\:\text{-}4 \\ u^2-u-11\:=\:0 & \Rightarrow & u \:=\:\dfrac{1\pm3\sqrt{5}}{2} \end{Bmatrix}$

Back-substitute and solve for $x.$
. . Be sure to check for extraneous roots.