Question

The value of $\sqrt{6+\sqrt{6+\sqrt{6+......\mathrm{to}\infty }}}$ is ____________.

Answer 1

$$\begin{gathered} \mathrm{Let}\sqrt{6+\sqrt{6+\sqrt{6+\_\_\_\_\_\infty }}}=x \\ =\sqrt{6+x}=x \\ \mathrm{i}.\mathrm{e}.6+x=x^2\left(\mathrm{squaring}\mathrm{both}\mathrm{sides}\right) \\ \mathrm{i}.\mathrm{e}.x^2-x-6=0 \\ \mathrm{i}.\mathrm{e}.x^2-3x+2x-6=0 \\ \mathrm{i}.\mathrm{e}.\left(x-3\right)\left(x+2\right)=0 \\ \mathrm{i}.\mathrm{e}.x=3\mathrm{or}x=-2 \\ \mathrm{Since}\mathit{}x=-2\mathrm{is}\mathrm{not}\mathrm{possible}(\mathrm{being}\mathrm{negative}) \\ \Rightarrow x=3 \\ \mathrm{i}.\mathrm{e}.\sqrt{6+\sqrt{6+\sqrt{6+\_\_\_\_\_\infty }}}=3 \end{gathered}$$