Question

what might be value would be​

Answer 1

Question :–

$\\ \sf \: If \: \: : \: \: x = \sqrt{3 + \sqrt{3 + \sqrt{3 + ....... \infty } } } \: \: \: \: then -$

$\\ \sf (1) \: \: {x}^{2} - x + 3 = 0$

$\\ \sf (3) \: \: {x}^{2} - x - 3 = 0$

ANSWER :–

GIVEN :–

$\\ \sf \:\: \: { \huge{.}} \: \: \: x = \sqrt{3 + \sqrt{3 + \sqrt{3 + ....... \infty } } }$

TO FIND :–

• 'x' in quadratic form = ?

SOLUTION :–

$\\ \sf \implies x = \sqrt{3 + \sqrt{3 + \sqrt{3 + ....... \infty } } }$

$\\ \sf \implies x = \sqrt{3 + \left[ \sqrt{3 + \sqrt{3 + ....... \infty } } \right] }$

• We should write this as –

$\\ \sf \implies x = \sqrt{3 + x }$

• square on both sides –

$\\ \sf \implies x {}^{2} = {3 + x }$

$\\ \sf \implies \large \boxed{ \sf \: x {}^{2} - x - 3 = 0}$

• Hence Option (3) is correct .

Answer 2

${\huge{\red{\sf{Given}}}}\begin{cases}\leadsto\bf{x=\sqrt{3+\sqrt{3+\sqrt{3+...\infty}}}} \end{cases}$

${\huge{\red{\sf{To\:Find}}}}\begin{cases}\leadsto\bf{(1)x^{2}-x+3}\\\bf{(3)x^{2}-x-3} \\\leadsto\bf{Correct\: option\: between\:them}\end{cases}$

$\huge\red{\underline{\bf{\green{Answer}}}}$

$\sf{\red{\hookrightarrow Taking\:the\:given\: equation,}}$

$\sf{\implies x =\sqrt{3+\sqrt{3+\sqrt{3+.....\infty}}}}$

$\sf{\implies x =\sqrt{3+x}}$

$\sf{\purple{Since\: x =\sqrt{3+\sqrt{3+\sqrt{3+...\infty}}}}}$

$\sf{\pink{\mapsto Squaring\:both\:sides}}$

$\sf{\implies x^{2}=3+x}$

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

${\underline{\boxed{\red{\bf{\therefore x^{2}-x-3=0}}}}}$

${\underline{\underline{\orange{\bf{Hence\:correct\: option\:is\:(3)}}}}}$