Question

x =
$x = \sqrt{6 + \sqrt{6 + \sqrt{ 6 \times \infty } } }$
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Answer 1

$x = \sqrt{6 + \sqrt{6 + \sqrt{ 6 \times \infty } } }$

The above expression is an infinite nested radical, since the equation is continuous. Substitute x in the radicals.

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

Squaring over both sides,

$\begin{gathered}\longrightarrow \sf x^2 = 6 + x \\ \\ \longrightarrow \sf x^2 -x - 6 = 0 \\ \\ \longrightarrow \sf x^2 - 3x + 2x -6 = 0 \\ \\ \longrightarrow \sf (x - 3)(x +2) = 0 \\ \\ \longrightarrow \sf x = 3 or -2\end{gathered}$

Since, x = -2 is beyond the scope of natural numbers, we shall ignore it.

$\longrightarrow \boxed{\boxed{\sf x = 3}}$

Answer 2

Answer:

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