JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 7)
Let $\alpha$ and $\beta$ be the roots of the equation $p x^2+q x-r=0$, where $p \neq 0$. If $p, q$ and $r$ be the consecutive terms of a non constant G.P. and $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$, then the value of $(\alpha-\beta)^2$ is :
8
9
$\frac{20}{3}$
$\frac{80}{9}$
Explanation
Given : $p x^2+q x-r=0$
Let $p=\frac{a}{r_1}, q=a, r=a r_1$
$\begin{aligned} & \text { and } \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} \\\\ & \Rightarrow \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \\\\ & \Rightarrow \frac{-\frac{q}{p}}{-\frac{r}{p}}=\frac{3}{4} \\\\ & \Rightarrow \frac{q}{r}=\frac{3}{4} \\\\ & \Rightarrow \frac{1}{r_1}=\frac{3}{4} \\\\ & \Rightarrow r_1=\frac{4}{3}\end{aligned}$
$\begin{aligned}(\alpha-\beta)^2 & =(\alpha+\beta)^2-4 \alpha \beta \\\\ & =\left(\frac{-q}{p}\right)^2-4\left(\frac{-r}{p}\right) \\\\ & =\frac{q^2}{p^2}+\frac{4 r}{p} \\\\ & =r_1^2+4 r_1^2=5 r_1^2 \\\\ & =5\left(\frac{4}{3}\right)^2=\frac{80}{9}\end{aligned}$
Let $p=\frac{a}{r_1}, q=a, r=a r_1$
$\begin{aligned} & \text { and } \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} \\\\ & \Rightarrow \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \\\\ & \Rightarrow \frac{-\frac{q}{p}}{-\frac{r}{p}}=\frac{3}{4} \\\\ & \Rightarrow \frac{q}{r}=\frac{3}{4} \\\\ & \Rightarrow \frac{1}{r_1}=\frac{3}{4} \\\\ & \Rightarrow r_1=\frac{4}{3}\end{aligned}$
$\begin{aligned}(\alpha-\beta)^2 & =(\alpha+\beta)^2-4 \alpha \beta \\\\ & =\left(\frac{-q}{p}\right)^2-4\left(\frac{-r}{p}\right) \\\\ & =\frac{q^2}{p^2}+\frac{4 r}{p} \\\\ & =r_1^2+4 r_1^2=5 r_1^2 \\\\ & =5\left(\frac{4}{3}\right)^2=\frac{80}{9}\end{aligned}$
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