Question:
The value of $\sqrt{6+\sqrt{6+\sqrt{6+\ldots \ldots \text { to } \infty}}}$ is
Solution:
Let $\sqrt{6+\sqrt{6+\sqrt{6+-----\infty}}}=x$
$=\sqrt{6+x}=x$
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. $\sqrt{6+\sqrt{6+\sqrt{6+_{-----\infty}}}}=3$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.