Question

If $\alpha$ and $\beta$ are the distinct roots of $a{x}^2+bx+c=0,$ where $a,b$ and $c$ are non-zero real numbers then
$\frac{a{\alpha }^2+b\alpha +6c}{a{\beta }^2+b{\beta }^2+9c}+\frac{a{\beta }^2+b{\beta }^2+19c}{a{\alpha }^2+b\alpha +13c}$ is equal to :

• A. $18c$
• B. $27c$
• C. $\frac{36}{27}$
• D. $\frac{17}{8}$
• E. $\frac{19}{13}$

Answer 1

The correct option is D $\frac{17}{8}$

Given equation is $a{x}^2+bx+c=0$

Since, $\alpha$ and $\beta$ are the roots of given equation.

Therefore, $a{\alpha }^2+b\alpha +c=0,a{\beta }^2+b\beta +c=0$

$\Rightarrow \frac{a{\alpha }^2+b\alpha +c+5c}{a{\beta }^2+b\beta +c+8c}+\frac{a{\beta }^2+b\beta +c+12c}{a{\alpha }^2+b\alpha +c+12c}$

$=\frac{0+5c}{8c}+\frac{0+18c}{0+12c}=\frac{5}{8}+\frac{18}{12}$

$=\frac{5}{8}+\frac{3}{2}$

$=\frac{5+12}{8}=\frac{17}{8}$