Question

2y² + 5y + 1 = 0

Answer 1

Given: 2y² + 5y + 1 = 0
=> 2y² + 5y = –1
On dividing both sides by 2, we get:
y² + $\frac{5}{2}$ y = $-\frac{1}{2}$ … (1)
We make L.H.S. a perfect square by using
Third term =${\left(\frac{1}{2}\mathrm{coefficient}\mathrm{of}y\right)}^2={\left(\frac{1}{2}\times \frac{5}{2}\right)}^2=\frac{25}{16}$
$$\begin{gathered} \mathrm{On}\mathrm{adding}\frac{25}{16}\text{to}\mathrm{both}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}: \\ {\mathrm{y}}^2+\frac{5}{2}\mathrm{y}+\frac{25}{16}=-\frac{1}{2}+\frac{25}{16} \\ =>y^2+2\times y\times \frac{5}{4}+{\left(\frac{5}{4}\right)}^2=\frac{17}{16} \\ =>{\left(y+\frac{5}{4}\right)}^2={\left(\frac{\sqrt{17}}{4}\right)}^2 \\ \text{Taking}\mathrm{squar}e\mathrm{root}\mathrm{of}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ y+\frac{5}{4}=\pm \frac{\sqrt{17}}{4} \\ =>y=-\frac{5}{4}\pm \frac{\sqrt{17}}{4}=\frac{-5\pm \sqrt{17}}{4} \\ \mathrm{Thus},y=\frac{-5+\sqrt{17}}{4},\frac{-5-\sqrt{17}}{4} \end{gathered}$$