Question

16x² -8x+1=0 when x=(1/4)y
,find the value of y​

Answer 1

Answer:

$firstly \: factorise \: it \\ 16 {x}^{2} - 8x + 1 = 0 \\ < by \: using \: identity > \\ < {a}^{2} - 2ab + {b}^{2} = {(a - b)}^{2} > \\ < where \: a \: = 4\: \: and \: b = 1 > \\ {(4x)}^{2} - 2 \times 4 \times 1 + {1}^{2} = 0 \\ (4x - 1) {}^{2} = 0 \\ 4x - 1 = \sqrt{0} = 0 \\ 4x = 1 \\ x = + \frac{1}{4} \: or - \frac{1}{4} \\ in \: question \: given \: that \\ x = \frac{y}{4} \\ then \\ y = 4x \\ = 4 \times \frac{1}{4} = 1 \\ or \: - 1 \\ oo \\ - \\ 0$