Question

0%2B%20%20%5Csqrt%7B6%7D%20%2B%20..........%20%20%5C%3A%20to%20%5C%3A%20%20%5Cinfin%20%7D%20%20%7D%20%20%5C%3A%20is%20%5C%3A%20%20%5Cunderline%7B%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20.%7D" id="TexFormula1" title="The \: value \: of \: \sqrt{6 + \sqrt{6 + \sqrt{6} + .......... \: to \: \infin } } \: is \: \underline{ \: \: \: \: \: \: \: \: \: \: \: .}" alt="The \: value \: of \: \sqrt{6 + \sqrt{6 + \sqrt{6} + .......... \: to \: \infin } } \: is \: \underline{ \: \: \: \: \: \: \: \: \: \: \: .}" align="absmiddle" class="latex-formula">
CORRECT EXPLANATION WILL BE MARK AS BRAINLIEST.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us assume that:

$\sf \longmapsto x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty } } }$

We can also write it as:

$\sf \longmapsto x = \sqrt{6 + x }$

Squaring both sides, we get:

$\sf \longmapsto {x}^{2} =6 + x$

$\sf \longmapsto {x}^{2} - x - 6 =0$

By splitting the middle term:

$\sf \longmapsto {x}^{2} - 3x + 2x - 6 =0$

$\sf \longmapsto x(x - 3) + 2(x - 3 ) =0$

$\sf \longmapsto (x+ 2)(x - 3 ) =0$

Therefore:

$\longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}$

$\sf \longmapsto x = - 2,3$

But x cannot be negative.

$\sf \longmapsto x = 3$

Therefore, the value of the expression is 3.

$\large\underline{\sf{Verification-}}$

Given:

$\sf\longmapsto x=3$

We can also write it as:

$\sf\longmapsto x = \sqrt{9}$

$\sf\longmapsto x = \sqrt{6+3}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{9}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}$

This pattern will continue.

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}$