Jee Main 2024 01 02 2024 Shift 2 - Question6

Answer: (1)

Solution:

$p x^{2}+q x-r=0$

Let $p=\frac{a}{r _1}, q=a, r=a r _1$

and $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$

$\frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4}$

$\Rightarrow \frac{-\frac{q}{p}}{-\frac{r}{p}}=\frac{3}{4}$

$\Rightarrow \frac{q}{r}=\frac{3}{4}$

$\Rightarrow \frac{1}{r _1}=\frac{3}{4}$

$\Rightarrow r _1=\frac{4}{3}$

$(\alpha-\beta)^{2}=\alpha^{2}+\beta^{2}-2 \alpha \beta$

$=(\alpha+\beta)^{2}-4 \alpha \beta$

$=\left(\frac{-q}{p}\right)^{2}-4\left(\frac{-r}{p}\right)$

$=\frac{q^{2}}{p^{2}}+\frac{4 r}{p}$

$$ \begin{aligned} & =r _1^{2}+4 r _1^{2} \\ & =5 r _1^{2} \\ & =5\left(\frac{4}{3}\right)^{2}=\frac{80}{9} \end{aligned} $$