How do I solve $\sqrt{x-\sqrt{x-\sqrt{x...}}}=9 \Rightarrow x=?$

Question

$\sqrt{x-\sqrt{x-\sqrt{x...}}}=9 \Rightarrow x=?$

I'm preparing for exam.

This question comes from metropol mat1 testbook.

Answer 1

Call $S=\sqrt{x-\sqrt{x-\sqrt{x...}}}$.

Square both sides:

$$x-\sqrt{x-\sqrt{x...}}=81$$ i.e. $$ x-S=81 $$ Now $S=9$, so $x=90$.

Answer 2

hint

Put $u_1=\sqrt{x}$ and for $ n>0$,

$$u_{n+1}=\sqrt{x-u_n}$$

for example

$$u_2=\sqrt{x-\sqrt{x}}$$ $$u_3=\sqrt{x-\sqrt{x-\sqrt{x}}}.$$

If this sequence converges, its limit should satisfy

$$L=\sqrt{x-L}$$ or

$$L^2+L-x=0$$ and $$L=\frac{-1+\sqrt{1+4x}}{2}=9$$

thus $$4x=(19+1)(19-1)$$ finally $$x=90$$