Question

Find the roots of the following quadratic equations by factorisation:
$2x^2-x+\frac{1}{8}=0$

Answer 1

$2x^2-x+\frac{1}{8}=\frac{1}{8}(16x^2-8x+1)$
(Splitting the middle term -8x as -4x-4x because $-4x\times -4x=16x^2$)

$=\frac{1}{8}(16x^2-4x-4x+1)=\frac{1}{8}\left(4x\right(4x-1)-1(4x-1\left)\right)=\frac{1}{8}(4x-1{)}^2$
Roots of this equation are the values for which $(4x-1{)}^2=0$
$\therefore (4x-1)=0$
$\Rightarrow x=\frac{1}{4}$
So both the roots of the equation is $\frac{1}{4}$