1. ## Linear Equation Q2

Question
$\displaystyle \frac{4}{25w^2}+\frac{3}{5w-1}=\frac{2}{5w-1}$

Attempt:

$\displaystyle \frac{4}{(5w+1)(5w-1)}+\frac{3}{5w-1}=\frac{2}{5w-1}$

$\displaystyle w=\frac{3}{20}$

Thank you

2. $\displaystyle \frac{4}{25w^2}+\frac{3}{5w-1}=\frac{2}{5w-1}$

$\displaystyle \frac{4}{25w^2}=\frac{-3}{5w-1}+\frac{2}{5w-1}$

$\displaystyle \frac{4}{25w^2}=\frac{-1}{5w-1}$

Hence

$\displaystyle 4(5w-1)= -25w^2$

$\displaystyle 20w-4=-25w^2$

$\displaystyle 25w^2+20w-4=0$
____________for next steps read the last lines if you don't get it_______

$\displaystyle w=\frac{-20\pm \sqrt{20^2-(4\times 25\times -4)}}{2\times 25}$

$\displaystyle w=\frac{-20\pm \sqrt{400+400}}{50}$

$\displaystyle w=\frac{20(-1+ \sqrt{2})}{50}~And~\frac{20(-1- \sqrt{2})}{50}$

$\displaystyle w=\frac{2( \sqrt{2}-1)}{5} ~And~ \frac{-2(1+\sqrt{2})}{5}$
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The two roots are $\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}$