Question

3y² + 7y + 1 =0

Answer 1

Given: 3y² + 7y + 1 =0

$\Rightarrow$ 3y² + 7y = –1

On dividing both sides by 3, we get:

$y^2+\frac{7}{3}y=\frac{-1}{3}$ ---(1)

We make L.H.S. a perfect square by using

Third term = .${\left(\frac{1}{2}\times \mathrm{coefficient}\mathrm{of}y\right)}^2={\left(\frac{1}{2}\times \frac{7}{3}\right)}^2={\left(\frac{7}{6}\right)}^2=\frac{49}{36}$

On adding $\frac{49}{36}$ to both sides of equation (1), we get:

$$\begin{gathered} y^2+\frac{7}{3}y+\frac{49}{36}=\frac{-1}{3}+\frac{49}{36} \\ \Rightarrow y^2+2\times y\times \frac{7}{6}+{\left(\frac{7}{6}\right)}^2=\frac{-12+49}{36}=\frac{37}{36} \\ \Rightarrow {\left(y+\frac{7}{6}\right)}^2={\left(\frac{\sqrt{37}}{6}\right)}^2 \end{gathered}$$

On taking square root of both sides, we get:

$$\begin{gathered} y+\frac{7}{6}=\pm \frac{\sqrt{37}}{6} \\ \Rightarrow y=-\frac{7}{6}\pm \frac{\sqrt{37}}{6}=\frac{-7\pm \sqrt{37}}{6} \end{gathered}$$

Thus, $y=\frac{-7\pm \sqrt{37}}{6}$.