Question

Please answer the question.
Note: Answer is multi-correct ​

Answer 1

substitute value of x

$x = \alpha \\ 4( \alpha ) {}^{2} + 2( \alpha ) - 1 = 0 \\ 4 \alpha {}^{2} + 2 \alpha - 1 = 0 \\ compring \: it \: with \: \\ \blue{ax {}^{2} + bx + c = 0} \\ a = 4 \\ b = 2 \\c = - 1 \\ \alpha = \frac{- b + - \sqrt{b {}^{2} - 4ac} } {2 a}$

$\frac{ - 2 + - \sqrt{2 {}^{2} - 4(4)( - 1)} }{2(4)}$

$= \frac{ - 2 + - \sqrt{4 + 16} }{8}$

$x = \frac{ - 2 + - \sqrt{20}}{8}$

$= \frac{ - 2 + - 2 \sqrt{5} }{8}$

$= \frac{ - 2 + 2 \sqrt{5} }{8} or \frac{ - 2 - 2 \sqrt{5} }{8}$

$x = \frac{ 2( \sqrt{5} - 1)}{8}or \: \frac{ - 2( \sqrt{5 } + 1) }{8}$

$x = \frac{ \sqrt{5 } - 1 }{4} or \frac{ \sqrt{5 } + 1}{4}$

use any one of these as

$\alpha \: and \: substitute \: in \\ all \: four \: options \: if \: you \: get \: \\ \alpha and \: the \: other \: value \: same \: thtoption \: \\ is \: correct \: \\ \green{leaving \: calculation \: part \: to \: you}$