Question

If $x>0$, $x=\sqrt{2}$ and $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}$, then the value of $2y-1$ is

• A. $\sqrt{2+4\sqrt{2}}$
• B. $\sqrt{2+4\sqrt{3}}$
• C. $\sqrt{1+4\sqrt{3}}$
• D. $\sqrt{1+4\sqrt{2}}$

Answer 1

The correct option is D $\sqrt{1+4\sqrt{2}}$

$y=\sqrt{x+\sqrt{x+\sqrt{x+....}}}=\sqrt{x+y}$

$\Rightarrow y^2-y=x$

Given $x=\sqrt{2}$

Therefore we get $y^2-y-\sqrt{2}=0$

We have a formula for solving quadratic equation $a{x}^2+bx+c=0$ is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow y=\frac{1+\sqrt{1+4\sqrt{2}}}{2}$

$\Rightarrow 2y-1=\sqrt{1+4\sqrt{2}}$