Question

2x^2 - x + 1/8 by completing the square​

Answer 1

Solution:-

Given equation is;

$2x {}^{2} - x + \frac{1}{8 } = 0$

Hence,

$16 {}^{2} - 8x + 1 = 0$

$16 {x}^{2} - 4x - 4x + 1 = 0$

$4x(4x - 1) - 1(4x - 1) = 0$

$(4x - 1) {}^{2} = 0$

$x = \frac{1}{4} \: and \: x = \frac{1}{4}$

Hence proved:)

Answer 2

Answer:

$Given equation is 2x ² −x+ \frac{1}{8} = 0$

16x ² −8x+1=0

16x ² −4x−4x+1=0

4x(4x−1)−1(4x−1)=0

(4x−1) ²=0

$x = \frac{1}{4} \: and \: \frac{1}{4}$

Hence proved :-)