Question

If $\alpha$ and $\beta$ are the roots of the equation $5x^2-2x+1=0,$ then the value of $\frac{{\alpha }^2}{1+\alpha }+\frac{{\beta }^2}{1+\beta }$ is equal to

• A. $\frac{1}{10}$
• B. $\frac{2}{5}$
• C. $-\frac{1}{10}$
• D. $-\frac{1}{5}$

Answer 1

The correct option is C $-\frac{1}{10}$

$5x^2-2x+1=0$
$\alpha +\beta =\frac{2}{5},\alpha \beta =\frac{1}{5}$

Now, $\frac{{\alpha }^2}{1+\alpha }+\frac{{\beta }^2}{1+\beta }$
$=\frac{{\alpha }^2(1+\beta )+{\beta }^2(1+\alpha )}{(1+\alpha )(1+\beta )}$
$=\frac{({\alpha }^2+{\beta }^2)+\alpha \beta (\alpha +\beta )}{1+(\alpha +\beta )+\alpha \beta }$
$=\frac{{\left(\frac{2}{5}\right)}^2-\frac{2}{5}+\frac{1}{5}\times \frac{2}{5}}{1+\frac{2}{5}+\frac{1}{5}}$
$=\frac{-1}{10}$