Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2+px+q=0$, then $-{\alpha }^{-1}$ and $-{\beta }^{-1}$ are the roots of which one of the following equations:

• A. $q{x}^2-px+1=0$
• B. $q^2+px+1=0$
• C. $x^2+px-q=0$
• D. $x^2-px+q=0$

Answer 1

The correct option is A

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

Explanation for the correct option:

Find the quadratic equation:

Since, $\alpha$ and $\beta$ are the roots of the equation $x^2+px+q=0$, then

Sum of root$=-\frac{\mathrm{coefficient}\mathrm{of}x}{\mathrm{coefficient}\mathrm{of}{x}^2}$

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

And

Product of root$=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{x}^2}$

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

By the given condition, $-{\alpha }^{-1}$ and $-{\beta }^{-1}$ are the roots of the required quadratic equation.

The required quadratic equation is $x^2-\left(\mathrm{sum}\mathrm{of}\mathrm{roots}\right)x+\left(\mathrm{product}\mathrm{of}\mathrm{roots}\right)=0$

Now,

Sum of roots $=-{\alpha }^{-1}-{\beta }^{-1}$

$\begin{array}{rcl}& \Rightarrow & -{\alpha }^{-1}-{\beta }^{-1}=-\left(\frac{1}{\alpha }+\frac{1}{\beta }\right)\\ & =& -\left(\frac{\beta +\alpha }{\alpha ·\beta }\right)\\ & =& -\left(\frac{-p}{q}\right)\\ & =& \frac{p}{q}\end{array}$

And

Product of roots$=-{\alpha }^{-1}·-{\beta }^{-1}$

$\begin{array}{rcl}& =& {\alpha }^{-1}·{\beta }^{-1}\\ & =& \frac{1}{\alpha \beta }\\ & =& \frac{1}{q}\end{array}$

Substitute these values in the required equation, we get

$$\begin{gathered} \Rightarrow x^2-\frac{p}{q}x+\frac{1}{q}=0 \\ \Rightarrow q{x}^2-px+1=0 \end{gathered}$$

Hence, the correct option is A.