Question

If $\alpha$ and $\beta$ are the equation, $7x^2-3x-2=0$, then the value of $\frac{\alpha }{1-{\alpha }^2}+\frac{\beta }{1-{\beta }^2}$ is equal to :

• A. $\frac{27}{32}$
• B. $\frac{1}{24}$
• C. $\frac{27}{16}$
• D. $\frac{3}{8}$

Answer 1

The correct option is C $\frac{27}{16}$

$\alpha +\beta =\frac{3}{7},\cdots \left(i\right)\alpha \beta =\frac{-2}{7}\cdots \left(ii\right)$
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$
$\Rightarrow {\alpha }^2+{\beta }^2=\frac{9}{49}+\frac{4}{7}\cdots \left(iii\right)$

$\frac{\alpha }{1-{\alpha }^2}+\frac{\beta }{1-{\beta }^2}$
$=\frac{(\alpha +\beta )-\alpha \beta (\alpha +\beta )}{1-({\alpha }^2+{\beta }^2)+(\alpha \beta {)}^2}$
$=\frac{\frac{3}{7}+\frac{2}{7}\times \frac{3}{7}}{1-\{\frac{9}{49}+\frac{4}{7}\}+\frac{4}{49}}$
$=\frac{\left(\frac{21+6}{49}\right)}{\frac{49-37+4}{49}}=\frac{27}{16}$
$\therefore \frac{(\alpha +\beta )-\alpha \beta (\alpha +\beta )}{1-({\alpha }^2+{\beta }^2)+(\alpha \beta {)}^2}=\frac{27}{16}$