Let α and β be the roots of the equation px 2 + qx - r =0 , where p neq 0 . If p , q and r be the consecutive terms of a non-constant G.P and 1/α+1/β=3/4 , then the value of (α-β) 2 is :

Mathematics · Quadratic Equations

JEE Main 2024 — 1 February, Shift 2 — Question 2

Let $\alpha$ and $\beta$ be the roots of the equation $\mathrm{px}^{2}+\mathrm{qx}-$ $\mathrm{r}=0$ , where $\mathrm{p} \neq 0$ . If $\mathrm{p}, \mathrm{q}$ 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 :

Option A:
$\frac{80}{9}$
Correct
Option B:
9
Option C:
$\frac{20}{3}$
Option D:
8

Answer: A

Step-by-step solution

$\mathrm{px}^{2}+\mathrm{qx}-\mathrm{r}=0<\beta$

$\mathrm{p}=\mathrm{A}, \mathrm{q}=\mathrm{AR},$

$\mathrm{r}=\mathrm{AR}^{2}$

$\mathrm{Ax}^{2}+\mathrm{ARx}-\mathrm{AR}^{2}=0$

$\mathrm{x}^{2}+\mathrm{Rx}-\mathrm{R}^{2}=0<_{\beta}^{\alpha}$

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

$\therefore \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \Rightarrow \frac{-\mathrm{R}}{-\mathrm{R}^{2}}=\frac{3}{4} \Rightarrow \mathrm{R}=\frac{4}{3}$

$(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=R^{2}-4\left(-R^{2}\right)=5\left(\frac{16}{9}\right)$

$=80 / 9$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Theory of Quadratic Equations