Question

Find the roots of the following equation by method of factorisation where p(x ) = 2x^2-x+1/8=0​

Answer 1

$\scriptsize\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}$

$\rule{300pt}{0.1pt}$

$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

$\rule{300pt}{0.1pt}$

Given equation is $\rm 2 x^{2}-x+\frac{1}{8}=0$

Multiplying both sides by 8 , we get

$\[ \begin{array}{l} \rm \Rightarrow 16 x^{2}-8 x+1=0 \\ \\ \rm \Rightarrow 16 x^{2}-(4 x+4 x)+1=0 \\ \\ \rm\Rightarrow 16 x^{2}-4 x-4 x+1=0 \\ \\ \rm \Rightarrow(4 x-1)-1(4 x-1)=0 \\ \\ \rm \Rightarrow(4 x-1)(4 x-1)=0 \end{array} \]$

$\text{Now, \( \rm \quad 4 x-1=0 \Rightarrow x=\dfrac{1}{4} \)}$

Hence, the roots of the equation $\rm 2 x^{2}- x+\dfrac{1}{8}=0$ are $\rm \dfrac{1}{4}$ and $\rm \dfrac{1}{4}$.