Mathematics - Quadratic Equation Question with Solution | TestHub

$(\alpha +\beta )=1$, $\alpha \beta =p$, $\gamma +\delta =4$, $\gamma \delta =q$. Since $\alpha ,\beta ,\gamma ,\delta$ are in G.P.,

$$\therefore \frac{\beta }{\alpha }=\frac{\delta }{\gamma }\Rightarrow \frac{\beta +\alpha }{\beta -\alpha }=\frac{\delta +\gamma }{\delta -\gamma }\Rightarrow \frac{(\beta +\alpha {)}^2}{(\beta +\alpha {)}^2-4\alpha \beta }=\frac{(\delta +\gamma {)}^2}{(\delta +\gamma {)}^2-4\delta \gamma }$$

$$\Rightarrow \frac{1}{1-4p}=\frac{16}{16-4q}=\frac{4}{4-q}$$

$$\Rightarrow 4-q=4-16p$$

Now, $p\ge 2\therefore q\ge 32$. For the given equation $a{x}^2+bx+c=0$ to have real roots, $b^2-4ac\ge 0$.

$$\therefore ac\le \frac{b^2}{4}$$

Hence, the number of quadratic equations with real roots, $r=24$.

Now from (i) and (ii), the minimum value of $pqr=2\cdot 32\cdot 24=1536$.