Question

If $\alpha ,\beta$ are the roots of $3X^2+5X-7=0$, then the value of ${\left(\frac{1}{3\alpha +5}\right)}^2+{\left(\frac{1}{3\beta +5}\right)}^2$ is

• A. $\frac{67}{63}$
• B. $\frac{67}{441}$
• C. $\frac{109}{441}$
• D. $\frac{109}{63}$

Answer 1

The correct option is B $\frac{67}{441}$

Given equation is $3X^2+5X-7=0$

Here, $a=3$, $b=5$ and $c=-7$

$\alpha$ and $\beta$ will be $\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-5\pm \sqrt{109}}{6}$

So, ${\left(\frac{1}{3\alpha +5}\right)}^2=\frac{4}{(5+\sqrt{109}{)}^2}$

${\left(\frac{1}{3\beta +5}\right)}^2=\frac{4}{(\sqrt{109}-5{)}^2}$

After Adding above two function, we get

$4(\frac{1}{(5+\sqrt{109}{)}^2}+\frac{1}{(\sqrt{109}-5{)}^2})=4\times \frac{2\times (109+25)}{(109-25{)}^2}=\frac{67}{441}$