Question

If $\alpha$ and $\beta$ are the zeros of the polynomial $f\left(x\right)=16x^2+4x-5$ then $\frac{1}{\alpha }+\frac{1}{\beta }$ is equal to

• A. $\frac{2}{5}$
• B. $\frac{5}{2}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is D $\frac{4}{5}$

If $\alpha$, $\beta$ are the zeros of the quadratic polynomial $f\left(x\right)=16x^2+4x-5$, then,
$(\alpha +\beta )=\frac{-b}{a}=-\frac{4}{16}=-\frac{1}{4},\alpha \beta =\frac{c}{a}=\frac{-5}{16}$

Now, $\frac{1}{\alpha }+\frac{1}{\beta }$ $=\frac{\alpha +\beta }{\alpha \beta }$

$=\frac{-\frac{1}{4}}{\frac{-5}{16}}$ $=\frac{16}{4\times 5}$

$=\frac{4}{5}$