Question

Let $\alpha ,\beta$ be the roots of $a{x}^2+bx+c=0$ and $\gamma ,\delta$ be the roots of $p{x}^2+qx+r=0$. If $\alpha ,\beta ,\gamma ,\delta$ are in H.P., then

• A. $\frac{b^2-4ac}{q^2-4pr}=\frac{c^2}{r^2}$
• B. $\frac{\alpha -\beta }{\alpha \beta }=\frac{1}{4}(\frac{b}{c}-\frac{q}{r})$
• C. $\frac{\gamma -\delta }{\gamma \delta }=\frac{1}{4}(\frac{b}{c}-\frac{q}{r})$
• D. $\frac{b^2-4ac}{q^2-4pr}=\frac{a^2}{p^2}$

Answer 1

The correct option is A $\frac{b^2-4ac}{q^2-4pr}=\frac{c^2}{r^2}$

$\alpha ,\beta ,\gamma ,\delta$ are in $HP$

$\Rightarrow \frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma },\frac{1}{\delta }$ are in $AP$.

$\alpha +\beta =\frac{-b}{a}$ and $\alpha \beta =\frac{c}{a}$

So, $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-b}{c}$ and $\frac{1}{\alpha }.\frac{1}{\beta }=\frac{a}{c}$

The quadratic equation with roots $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ is $c{x}^2+bx+a$.

Similarly, the quadratic equation with roots $\frac{1}{\gamma }$ and $\frac{1}{\delta }$ is $r{x}^2+qx+p$.

$\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma },\frac{1}{\delta }$ are in AP.

$\Rightarrow |\frac{1}{\alpha }-\frac{1}{\beta }|=|\frac{1}{\gamma }-\frac{1}{\delta }|$

$\Rightarrow \frac{\sqrt{b^2-4ac}}{|c|}=\frac{\sqrt{q^2-4pr}}{|r|}$

Squaring both sides, we get

$\frac{b^2-4ac}{c^2}=\frac{q^2-4pr}{r^2}$

$\Rightarrow \frac{b^2-4ac}{q^2-4pr}=\frac{c^2}{r^2}$