Question

Plz Solve this Question From Equation Chapter..Plz Help​

Answer 1

$\huge\mathfrak\red{Answer}$

$let \\ 4 + \frac{1}{4 + \frac{1}{4 + \frac{1}{4 + \frac{1}{.......infinite} } } } = x.....(1) \\ \\ x = 4 + \frac{1}{x} \\ as \: you \: see \: the \: whole \: equation \\ in \: denominator \: is \: similiar \: to \: \\ what \: we \: assume \\ x = 4 + \frac{1}{x} \\ {x}^{2} = {4x + 1} \\ \\ solve \: for \: x \\ \\ {x}^{2} - 4x - 1 = 0 \\ \\ a = 1 \\ b = - 4 \\ c = - 1 \\ by \: using \: quadratic \: formula \\ \\ x = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \: or \: \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x = \frac{4 + \sqrt{16 - 4(1)( - 1)} }{2(1)} \\ \\ x = \frac{4 + \sqrt{16 + 4} }{2} \\ \\ x = \frac{4 + \sqrt{20} }{2} \\ \\ x = \frac{4 + 2 \sqrt{5} }{2} \\ \\ x =2 + 2 \sqrt{5} \: or \: 2 - 2 \sqrt{5} \\ \\ so \: from \: eq(1) \\ \\ 4 + \frac{1}{4 + \frac{1}{4 + \frac{1}{4 + \frac{1}{ ......infinite} } } } = x = 2 + 2 \sqrt{5} \: or \: 2 - 2 \sqrt{5}$

OPTION C