# Prove the following

Question:

If $\alpha$ and $\beta$ are the roots of the equation, $7 x^{2}-3 x-2=0$,

the the value of $\frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$ is equal to :

1. (1) $\frac{27}{32}$

2. (2) $\frac{1}{24}$

3. (3) $\frac{3}{8}$

4. (4) $\frac{27}{16}$

Correct Option: , 4

Solution:

Let $\alpha$ and $\beta$ be the roots of the quadratic equation

$7 x^{2}-3 x-2=0$

$\therefore \alpha+\beta=\frac{3}{7}, \alpha \beta=\frac{-2}{7}$

Now, $\frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$

$=\frac{\alpha-\alpha \beta(\alpha+\beta)+\beta}{1-\left(\alpha^{2}+\beta^{2}\right)+(\alpha \beta)^{2}}$

$=\frac{(\alpha+\beta)-\alpha \beta(\alpha+\beta)}{1-(\alpha+\beta)^{2}+2 \alpha \beta+(\alpha \beta)^{2}}$

$=\frac{\frac{3}{7}+\frac{2}{7} \times \frac{3}{7}}{1-\frac{9}{49}+2 \times \frac{-2}{7}+\frac{4}{49}}=\frac{27}{16}$