Question

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

$\frac{27}{16}$

Explanation for the correct option:

Step 1: Find the sum and product of the roots of the given quadratic equation.

We have given a quadratic equation$7x^2-3x-2=0$, whose roots are$\alpha$ and$\beta$.

and we have to find the value of$\frac{\alpha }{1-{\alpha }^2}$ and$\frac{\beta }{1-{\beta }^2}$.

So,

$\begin{array}{rcl}\alpha +\beta & =& \frac{-b}{a}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\alpha +\beta & =& \frac{-(-3)}{7}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\alpha +\beta & =& \frac{3}{7}\end{array}$

$\begin{array}{rcl}\alpha \beta & =& \frac{c}{a}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\alpha \beta & =& \frac{-2}{7}\end{array}$

Step 2: Simplify the expression$\frac{\mathrm{\alpha }}{1-{\mathrm{\alpha }}^2}$ and$\frac{\mathrm{\beta }}{1-{\mathrm{\beta }}^2}$.

$\begin{array}{rcl}\frac{\alpha }{1-{\alpha }^2}+\frac{\beta }{1-{\beta }^2}& =& \frac{\alpha (1-{\beta }^2)+\beta (1-{\alpha }^2)}{(1-{\alpha }^2)(1-{\beta }^2)}\\ & =& \frac{\alpha -\alpha {\beta }^2+\beta -{\alpha }^2\beta }{1-{\beta }^2-{\alpha }^2+{\alpha }^2{\beta }^2}\\ & =& \frac{(\alpha +\beta )-\alpha \beta (\beta +\alpha )}{1-[{\alpha }^2+{\beta }^2]+{\left(\alpha \beta \right)}^2}\\ & =& \frac{(\alpha +\beta )[1-\alpha \beta ]}{1-[{(\alpha +\beta )}^2-2\alpha \beta ]+{\left(\alpha \beta \right)}^2}\end{array}$

Step 3: Substitute the value in the above expression.

$$\begin{gathered} \frac{(\alpha +\beta )[1-\alpha \beta ]}{1-[{(\alpha +\beta )}^2-2\alpha \beta ]+{\left(\alpha \beta \right)}^2} \\ =\frac{\left(\frac{3}{7}\right)\left[1-\left(-\frac{2}{7}\right)\right]}{1-\left[{{\displaystyle \left(\frac{3}{7}\right)}}^2-2\left(-\frac{2}{7}\right)\right]+{\left(-\frac{2}{7}\right)}^2} \\ =\frac{\left(\frac{{\displaystyle 3}}{{\displaystyle 7}}\right)\left(1+{\displaystyle \frac{{\displaystyle 2}}{{\displaystyle 7}}}\right)}{1-\left(\frac{9}{49}+{\displaystyle \frac{4}{7}}\right)+{\displaystyle \frac{{\displaystyle 4}}{{\displaystyle 49}}}} \\ =\frac{{\displaystyle \frac{{\displaystyle 27}}{{\displaystyle 49}}}}{1-{\displaystyle \frac{{\displaystyle 37}}{{\displaystyle 49}}}+{\displaystyle \frac{{\displaystyle 4}}{{\displaystyle 49}}}} \\ =\frac{{\displaystyle \frac{{\displaystyle 27}}{{\displaystyle 49}}}}{{\displaystyle \frac{{\displaystyle 16}}{{\displaystyle 49}}}} \\ =\frac{\mathbf{27}}{\mathbf{16}} \end{gathered}$$

Hence, the correct option is$\mathbf{\left(}\mathit{C}\mathbf{\right)}$.