Question

root 6 plus root 6 till infinity is equal to

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{are}\mathrm{given} \\ \sqrt{6+\sqrt{6+\sqrt{6+...........\infty }}} \\ \mathrm{let}\mathrm{x}=\sqrt{6+\sqrt{6+\sqrt{6+...........\infty }}} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ {\mathrm{x}}^2=6+\sqrt{6+\sqrt{6+\sqrt{6+...........\infty }}} \\ \Rightarrow {\mathrm{x}}^2=6+\mathrm{x}\left(\sqrt{6+\sqrt{6+\sqrt{6+...........\infty }}}=\mathrm{x}\right) \\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}-6=0 \\ \Rightarrow {\mathrm{x}}^2-3\mathrm{x}+2\mathrm{x}-6=0 \\ \Rightarrow \mathrm{x}\left(\mathrm{x}-3\right)+2\left(\mathrm{x}-3\right)=0 \\ \Rightarrow \left(\mathrm{x}-3\right)\left(\mathrm{x}+2\right)=0 \\ \Rightarrow \mathrm{x}=-2,3 \\ \mathrm{As},\sqrt{6+\sqrt{6+\sqrt{6+...........\infty }}}\mathrm{is}\mathrm{positive}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}\sqrt{6+\sqrt{6+\sqrt{6+...........\infty }}}\mathrm{is}3 \end{gathered}$$