Question

Let $\alpha, \beta$ be the roots of $x^2 + 3x + 5= 0$ then the equation whose roots are $ - \frac{1}{\alpha}$ and $ - \frac{1}{\beta}$ is .

• A. $5x^2 - 3x +1= 0$
• B. $5x^2 + 3x - 4= 0$
• C. $5x^2 - 3x + 4= 0$
• D. $5x^2 + 3x -1= 0$

Answer 1

Answer: (A)

Given $\alpha, \beta$ be the roots of equation
$x^{2}+3 x+5=0$
$\therefore \, \begin{cases}\alpha+\beta & =-3 \\ \alpha \beta & =5\end{cases}\,\,\,\,\,\dots(i)$
Now, the equation whose roots are $\frac{-1}{\alpha}$ and $\frac{-1}{\beta}$
will be
$x^{2}-\left(-\frac{1}{\alpha}-\frac{1}{\beta}\right) x+\left(-\frac{1}{\alpha}\right)\left(-\frac{1}{\beta}\right)=0 $
$x^{2}+\left(\frac{\alpha+\beta}{\alpha \beta}\right) x+\frac{1}{\alpha \beta}=0$
Now, from Eq. (i), we get
$ x^{2}+\left(-\frac{3}{5}\right) x+\frac{1}{5}=0 $
$\Rightarrow \, 5 x^{2}-3 x+1=0$