Question

$\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}}$ up to $\infty$ is?

• A. $2$
• B. $3$
• C. $30$
• D. $5$

Answer 1

The correct option is B $3$

We have,

$y=\sqrt{6+\sqrt{6+....\infty }}=\sqrt{6+y}$

On squaring both sides, we get

$\Rightarrow y^2-y-6=0$

$\Rightarrow y=3,-2$

But $y$ can not be $-ve$.

Hence, $y=3$ is the answer.