The value of


The value of $\sqrt{6+\sqrt{6+\sqrt{6+\ldots \ldots \text { to } \infty}}}$ is


Let $\sqrt{6+\sqrt{6+\sqrt{6+-----\infty}}}=x$


i. e. $6+x=x^{2}$ (squaring both sides)

i. e. $x^{2}-x-6=0$

i. e. $x^{2}-3 x+2 x-6=0$

i.e. $(x-3)(x+2)=0$

i.e. $x=3$ or $x=-2$

Since $x=-2$ is not possible (being negative)

$\Rightarrow x=3$

i. e. $\sqrt{6+\sqrt{6+\sqrt{6+_{-----\infty}}}}=3$

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