Question

what is the value of x​

Answer 1

x = $\sqrt{6+\sqrt{6+\sqrt{6...}}}$

Squaring both sides,

${x} ^{2}$ = 6 + $\sqrt{6+\sqrt{6+\sqrt{6...}}}$

${x} ^{2}$ = 6 + x

${x} ^{2}$ - x - 6 = 0

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

x ( x - 3 ) + 2 ( x - 3 ) = 0

( x - 3 ) ( x + 2 ) = 0

( x - 3) = 0, ( x +2 ) = 0

$\fbox{x \:=\: 3,\: - \:2}$