[tex]\large\underline{\sf{Solution-}}[/tex]
Let us assume that:
[tex] \sf \longmapsto x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty } } } [/tex]
We can also write it as:
[tex] \sf \longmapsto x = \sqrt{6 + x } [/tex]
Squaring both sides, we get:
[tex] \sf \longmapsto {x}^{2} =6 + x[/tex]
[tex] \sf \longmapsto {x}^{2} - x - 6 =0[/tex]
By splitting the middle term:
[tex] \sf \longmapsto {x}^{2} - 3x + 2x - 6 =0[/tex]
[tex] \sf \longmapsto x(x - 3) + 2(x - 3 ) =0[/tex]
[tex] \sf \longmapsto (x+ 2)(x - 3 ) =0[/tex]
Therefore:
[tex] \longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}[/tex]
[tex] \sf \longmapsto x = - 2,3[/tex]
But x cannot be negative.
[tex] \sf \longmapsto x = 3[/tex]
Therefore, the value of the expression is 3.
[tex]\large\underline{\sf{Verification-}}[/tex]
Given:
[tex]\sf\longmapsto x=3[/tex]
We can also write it as:
[tex]\sf\longmapsto x = \sqrt{9}[/tex]
[tex]\sf\longmapsto x = \sqrt{6+3}[/tex]
[tex]\sf\longmapsto x = \sqrt{6 + \sqrt{9}}[/tex]
[tex]\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}[/tex]
[tex]\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}[/tex]
[tex]\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}[/tex]
This pattern will continue.
[tex]\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}[/tex]
