Question

Solve the following quadratic equation by completing square method.

$3p^2+4=-7p$.

• A. $-1$
• B. $-\frac{4}{3}$
• C. $2$
• D. $3$

Answer 1

Answer: (A), (B)

The correct options are
A $-1$
B $-\frac{4}{3}$

The given equation is $3p^2+4=-7p$

$3p^2+4=-7p$

$\Rightarrow$ $3p^2+7p+4=0$

$\Rightarrow$ $p^2+\frac{7}{3}p+\frac{4}{3}=0$ [ Dividing both sides by $3$ ]

$\Rightarrow$ $p^2+\frac{7}{3}p=-\frac{4}{3}$

$\Rightarrow$ $p^2+2\times p\times \frac{7}{6}=-\frac{4}{3}$

$\Rightarrow$ $p^2+2\times p\times \frac{7}{6}+{\left(\frac{7}{6}\right)}^2=-\frac{4}{3}+{\left(\frac{7}{6}\right)}^2$

$\Rightarrow$ ${(p+\frac{7}{6})}^2=-\frac{4}{3}+\frac{49}{36}$

$\Rightarrow$ ${(p+\frac{7}{6})}^2=\frac{-48+49}{36}$

$\Rightarrow$ ${(p+\frac{7}{6})}^2=\frac{1}{36}$

$\Rightarrow$ ${(p+\frac{7}{6})}^2={\left(\frac{1}{6}\right)}^2$

$\Rightarrow$ $p+\frac{7}{6}=\pm \frac{1}{6}$

$\Rightarrow$ $p=\pm \frac{1}{6}-\frac{7}{6}$

$\Rightarrow$ $p=\frac{1}{6}-\frac{7}{6}$ or $p=\frac{-1}{6}-\frac{7}{6}$

$\Rightarrow$ $p=\frac{-6}{6}$ or $p=\frac{-8}{6}$

$\Rightarrow$ $p=-1$ or $p=\frac{-4}{3}$

Hence the solution is $-1$ and $\frac{-4}{3}$