Question

If the roots of the equation $p{x}^2+qx+r=0$, is 3:4 then $12q^2$= ______

• A. 49pr
• B. 36pr
• C. 64pr
• D. 81pr

Answer 1

The correct option is A 49pr

For the given equation, let the roots be $3a,4a$

So sum of roots $=3a+4a=7a=-\frac{q}{p}$

$=>a=-\frac{q}{7p}$

Product of roots $=3a\times 4a==12a^2=\frac{r}{p}$

$=>12{(-\frac{q}{7p})}^2=\frac{r}{p}$

$=>12\left(\frac{q^2}{49p^2}\right)=\frac{r}{p}$

$=>12q^2=49pr$