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
$x^2+3x+5=0$
$\therefore \{\begin{array}{ll}\alpha +\beta & =-3\\ \alpha \beta & =5\end{array}\dots (i)$
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
$x^2+\left(-\frac{3}{5}\right)x+\frac{1}{5}=0$
$\Rightarrow 5x^2-3x+1=0$