Question

(2x)^2-x+1/8=0 solve by factorisation method

Answer 1

Check the attachment.

Answer 2

Answer:

$x = \frac{1}[4},\frac{1}{4}$

Step-by-step explanation:

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

To Find: x

Solution:

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

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

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

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

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

$x = \frac{1}[4},\frac{1}{4}$

Hence the solution is $x = \frac{1}[4},\frac{1}{4}$