Question

$x+\sqrt{12-\sqrt{x}}=12$

Solve this equation for x.

Please provide the full steps.

Answer 1

$$0\le x\le12$$

$$\sqrt{12-\sqrt{x}}=12-x$$ $$\bigg(\sqrt{12-\sqrt{x}}\bigg)^2=(12-x)^2$$

$$12-\sqrt{x}=144-24x+x^2$$

$$x^2-24x+\sqrt{x}+132=0$$

$$(\sqrt{x})^4-24(\sqrt{x})^2+\sqrt{x}+132=0$$

$$(\sqrt{x}+4)(\sqrt{x}-3)((\sqrt{x})^2-\sqrt{x}-11)=0$$

$$\sqrt{x}+4>0$$

$$\sqrt{x}-3=0=>x=9$$

$$(\sqrt{x})^2-\sqrt{x}-11=0$$

$$D=(-1)^2-4(1)(-11)=45$$

$$\sqrt{x}=\dfrac{1\pm\sqrt{45}}{2}$$

Since $\sqrt{x}\ge0,$ we take

$$\sqrt{x}=\dfrac{1+\sqrt{45}}{2}$$

$$x=\dfrac{(1+3\sqrt{5})^2}{4}$$

$$x=\dfrac{23+3\sqrt{5}}{2}>12, does\ not\ satisfy$$

$\{9\}$