Question

If $\alpha ,\beta$ are the roots of the equation $x^2+px+q=0$, then $-\frac{1}{\alpha },-\frac{1}{\beta }$ are the roots of the equation

Answer 1

Given: $\alpha ,\beta$ are the roots of the equation $x^2+px+q=0$

Now,

$\text{sum of roots}=-\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}$

$\Rightarrow \alpha +\beta =-p\dots \left(1\right)$

$\text{Product of roots}=\frac{\text{constant term}}{\text{coefficient of}{x}^2}$

$\Rightarrow \alpha \beta =q\dots \left(2\right)$

The quadratic equation having roots

$-\frac{1}{\alpha },-\frac{1}{\beta }$ is given by:

$x^2-\text{(sum of roots )}x+\text{Product of roots}=0$

$\Rightarrow x^2-(-\frac{1}{\alpha }-\frac{1}{\beta })x+(-\frac{1}{\alpha }\times -\frac{1}{\beta })=0$

$\Rightarrow x^2+\left(\frac{\alpha +\beta }{\alpha \beta }\right)x+\left(\frac{1}{\alpha \beta }\right)=0$

From $\left(1\right)$ and $\left(2\right)$, we get

$\Rightarrow x^2+\left(\frac{-p}{q}\right)x+\left(\frac{1}{q}\right)=0$

$\therefore q{x}^2-px+1=0$

Hence, option (D) is correct.