Question

Find the value of $x$ if $x=\sqrt{6+\sqrt{6+\sqrt{+6+.....}}}$. Where $x$ is a natural number.

Answer 1

Compute the required value:

Given:

$x=\sqrt{6+\sqrt{6+\sqrt{6+.....}}}$

It can be written as:

$x=\sqrt{6+x}$

Squaring both sides and equating to zero

$$\begin{gathered} \Rightarrow x^2-x-6=0 \\ \Rightarrow x^2-3x+2x-6=0 \\ \Rightarrow x(x-3)+2(x-3)=0 \\ \Rightarrow \left(x-3\right)\left(x+2\right)=0 \\ \Rightarrow x=3,-2 \end{gathered}$$

Hence, the value of $x$ is $3$ as negative value is not considered under the square root.