Mathematics - Quadratic Equation Question with Solution | TestHub

Rewrite the given equation $\sqrt{x+2\sqrt{x+2\sqrt{x+\dots \dots \dots \dots ...}2\sqrt{x+2\sqrt{x+2x}}}}=x$ .......(1)

On replacing the last letter x on the LHS of eq.(1) by the value of x expressed by Eq.(1) we obtain

$x=\sqrt{x+2\sqrt{x+2\sqrt{x+\dots \dots \dots \dots +2\sqrt{x+2x}}}}$ (2n radicals)

further, let us replace the last letter x by the same expression

we can write $x=\sqrt{x+2\sqrt{x+2\sqrt{x+\dots \dots \dots \dots ...}}}$ $={\lim }_{N\to \infty }\sqrt{x+2\sqrt{x+2\sqrt{x+\dots \dots ..}+2\sqrt{x+2x}}}$

It follows that $x=\sqrt{x+2\sqrt{x+2\sqrt{x+\dots \dots \dots \dots .....}}}=\sqrt{x+2(\sqrt{x+2\sqrt{x+\dots \dots \dots ..}})}=\sqrt{x+2x}$ Hence $x=\sqrt{x+2x}\Rightarrow x^2=3x\Rightarrow x(x-3)=0$

$x=0$ (or) $x=3$

∴ Sum of roots $=3$