Question

Solve the quadratic equation, $2x^2-x+\frac{1}{8}=0$, using factorisation.

Answer 1

The given equation is $2x^2-x+\frac{1}{8}$. This can be written as,
$\frac{1}{8}(16x^2-8x+1)=0$ [0.5 Marks]

Now, to solve it by factorisation, split the middle term $-8x$ as $-4x-4x$ because $-4x\times -4x=16x^2$, we get

$\begin{array}{cc}\frac{1}{8}(16x^2-4x-4x+1)& =0\\ \frac{1}{8}\left(4x\right(4x-1)-1(4x-1\left)\right)& =0\\ \frac{1}{8}(4x-1{)}^2& =0\\ (4x-1{)}^2& =0\\ 4x-1& =0\\ x& =\frac{1}{4}\end{array}$
[1 Mark]

So, both the roots of the given equation is $\frac{1}{4}$. [0.5 Marks]