Question

5m² + m = 3

Answer 1

Given: 5m² + m = 3
On dividing both sides by 5, we get:
$m^2+\frac{1}{5}m=\frac{3}{5}$ …(1)
We make L.H.S. a perfect square by using,

$$\begin{gathered} \mathrm{Third}\mathrm{term}={\left(\frac{1}{2}\mathrm{coefficient}\mathrm{of}\mathrm{m}\right)}^2 \\ ={\left(\frac{1}{2}\times \frac{1}{5}\right)}^2={\left({\displaystyle \frac{1}{10}}\right)}^2=\frac{1}{100} \\ \mathrm{On}\mathrm{addding}\frac{1}{100}\mathrm{to}\mathrm{both}\mathrm{sides}\mathrm{of}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}: \\ {m}^2+\frac{1}{5}m+\frac{1}{100}=\frac{3}{5}+\frac{1}{100} \\ \Rightarrow m^2+2\times m\times \frac{1}{10}+{\left(\frac{1}{10}\right)}^2=\frac{61}{100} \\ \Rightarrow {\left(m+\frac{1}{10}\right)}^2=\frac{61}{100} \\ \text{T}\mathrm{aking}\mathrm{square}\mathrm{roots}\text{of}\mathrm{both}\mathrm{the}\mathrm{sides},\text{we get:} \\ \left(m+\frac{1}{10}\right)=\pm \frac{\sqrt{61}}{10} \\ \Rightarrow m=-\frac{1}{10}\pm \frac{\sqrt{61}}{10} \\ \Rightarrow m=\frac{-1+\sqrt{61}}{10},\frac{-1-\sqrt{61}}{10} \end{gathered}$$