Question

$solve \: x = \sqrt{6 + \sqrt{6 + \sqrt{6} } } .........$
​

Answer 1

$\huge\underline\mathbb{SOLUTION:-}$

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

$\implies \mathsf {x = \sqrt{6 + x}}$

$\implies \mathsf {x {}^{2} = 6 + x}$

$\implies \mathsf {x {}^{2} - x - 6 = 0}$

$\implies \mathsf {(x - 3)(x + 2) = 0}$

$\implies \mathsf \blue {x = 3}$

Answer 2

$\huge\bf{\underline{\red{Solution:-}}}$

Given:-

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

$\sf\:\:\:\:\:\:\:\:\:we\:have\:to\: find\:value\:of\:x$

$\bf\underline{\red{A.T.Q:-}}$

$\sf\implies x=\sqrt{6+\sqrt{6+\sqrt{6.....\infty}}}\\ \\ \sf\implies x=\sqrt{6+x}\\ \\ \sf\:\:\:\:\:\: squaring\:both\:side \\ \\ \sf\implies x{}^{2}=(\sqrt{6+x}){}^{2}\\ \\ \sf\implies x{}^{2}= 6+x\\ \\ \sf\implies x{}^{2}-x-6=0\:\:\:\:(quadratic\: equation)\\ \\ \sf\implies x{}^{2}-(3-2)x-6=0\\ \\ \sf\implies x{}^{2}-3x+2x-6=0\\ \\ \sf\implies x(x-3)+2(x-3)=0\\ \\ \sf\implies (x-3)(x+2)=0\\ \\ \sf\:\:\:\:\: \bullet x=3 \:\:and\:\:(-2)$

Now

$\sf\bullet \sqrt{6+\sqrt{6+\sqrt{6+.....\infty.}}} \neq (-2)\\ \sf\:\:\:\:\:\:\:\:\:(bcz.\:this\:is\: positive)\\ \\ \sf\bullet \sqrt{6+\sqrt{6+\sqrt{6+......\infty}}}= 3$

$\boxed{\sf{\purple{\therefore \sqrt{6+\sqrt{6+\sqrt{6.......\infty=3}}}}}}$