Question

Find the possible value(s) of the expression: $\sqrt{2+\sqrt{2+\sqrt{2+...}}}$

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

Answer: (B), (C)

$2$

Let $\sqrt{2+\sqrt{2+\sqrt{2+...}}}=x$

Substituting $x$ in the above expression,
$\Rightarrow \sqrt{2+x}=x$
(squaring both side of the equation)
$\Rightarrow 2+x=x^2$
$\Rightarrow x^2-x-2=0$
$\Rightarrow x^2-(2-1)x-2=0$
$\Rightarrow x^2-2x+x-2=0$
​​​​​​​$\Rightarrow x(x-2)+1\times (x-2)=0$
(taking $(x-2)$ as common term)
​​​​​​​$\Rightarrow (x-2)(x+1)=0$

$\therefore$ Either $x-2=0$ or $x+1=0.$

For $x-2=0$
$\Rightarrow x=2$

For $x+1=0$
​​​​​​​$\Rightarrow x=-1$

$\therefore$ Possible values of the given expression are $2$ and $-1.$