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Question in Latex: $\displaystyle x = 12-\sqrt{12-\sqrt{x}}$Quote:
Originally Posted by Mhmh96
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 $\displaystyle x \geq 0 \text{ and } \sqrt{x} \leq 12 \Leftrightarrow x \leq 144$. Overall this means $\displaystyle 0 \leq x \leq 144$
Edit: Don't forget to test potential solutions in the original equation to check if they are spurious or not
I tried to square both side but still not able to
determine to the right answer which is 9(Worried)
Hello, Mhmh96!
Quote:
$\displaystyle \text{Solve for }x\!:\;\;x \;=\;12 - \sqrt{12-\sqrt{x}}$
Let $\displaystyle u \,=\, \sqrt{x}\quad\Rightarrow\quad x \,=\,u^2$
We have: .$\displaystyle u^2 \:=\:12 - \sqrt{12-u} \quad\Rightarrow\quad \sqrt{12-u} \;=\;12 - u^2$
Square both sides: .$\displaystyle \left(\sqrt{12-u}\right)^2 \;=\; \left(12-u^2\right)^2$
. . $\displaystyle 12-u \;=\;144 - 24u^2 + u^4 \quad\Rightarrow\quad u^4 - 24u^2 + u + 132 \:=\:0$
which factors: .$\displaystyle (u-3)(u+4)(u^2-u-11) \:=\:0$
We have: .$\displaystyle \begin{Bmatrix} u-3 \:=\:0 & \Rightarrow & u \:=\:3 \\ \\[-3mm] u+4\:=\:0 & \Rightarrow & u \:=\:\text{-}4 \\ u^2-u-11\:=\:0 & \Rightarrow & u \:=\:\dfrac{1\pm3\sqrt{5}}{2} \end{Bmatrix}$
Back-substitute and solve for $\displaystyle x.$
. . Be sure to check for extraneous roots.
