Question

If the roots of $a{x}^2+bx+c=0$ and $p{x}^2+qx+r=0$ differ by the same quantity, then$\frac{b^2-4ac}{q^2-4pr}$ is

• A. ${\left(\frac{p}{a}\right)}^2$
• B. ${\left(\frac{c}{p}\right)}^2$
• C. ${\left(\frac{a}{p}\right)}^2$
• D. ${\left(\frac{p}{c}\right)}^2$

Answer 1

The correct option is C ${\left(\frac{a}{p}\right)}^2$

Let roots of $a{x}^2+bx+c=0$ be $\alpha$ and $\beta .$
Let roots of $p{x}^2+qx+r=0$ be $\sigma$ and $\delta .$
Then, $|\alpha -\beta |=|\sigma -\delta |$

$\therefore (\alpha -\beta {)}^2=(\sigma -\delta {)}^2$.
$(\alpha +\beta {)}^2-4\alpha \beta =(\sigma +\delta {)}^2-4\sigma \delta$ ........(1)

But $\alpha +\beta =\frac{-b}{a}$ $\sigma +\delta =\frac{-q}{p}$

$\alpha \beta =\frac{c}{a}$ $\sigma \delta =\frac{r}{p}$

Substituting values in (1) we get
$\frac{b^2-4ac}{a^2}=\frac{q^2-4pr}{p^2}$
$\Rightarrow \frac{b^2-4ac}{q^2-4pr}={\left(\frac{a}{p}\right)}^2$