Question
Find the roots of the quadratic equation $2 x^2-x+\frac{1}{8}=0$ by factorisation method.
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Answer
$2 x^2-x+\frac{1}{8}=0$
$\therefore 16 x^2-8 x+1=0 \ ($Multiplying by $8)$
$\therefore 16 x^2-4 x-4 x+1=0$
$\therefore 4 x(4 x-1)-1(4 x-1)=0$
$\therefore(4 x-1)(4 x-1)=0$
$\therefore 4 x-1=0 \text { or } 4 x-1=0$
$\therefore x=\frac{1}{4}$ or $ x=\frac{1}{4}$
Thus, the required roots of the given equation are $\frac{1}{4}$ and $\frac{1}{4}$.
$\therefore 16 x^2-8 x+1=0 \ ($Multiplying by $8)$
$\therefore 16 x^2-4 x-4 x+1=0$
$\therefore 4 x(4 x-1)-1(4 x-1)=0$
$\therefore(4 x-1)(4 x-1)=0$
$\therefore 4 x-1=0 \text { or } 4 x-1=0$
$\therefore x=\frac{1}{4}$ or $ x=\frac{1}{4}$
Thus, the required roots of the given equation are $\frac{1}{4}$ and $\frac{1}{4}$.
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