Let α ,β be the roots of the equation x 2 - 3 x + r = 0 and α/2, 2β be the roots of the equation x 2 + 3 x + r = 0 . If the roots of the equation x 2 + 6 x = m are 2α +β +2 r and α - 2β -frac r 2 , then m is equal to :-

Mathematics · Quadratic Equations

JEE Main 2026 — 2 April, Evening Shift — Question 23

Let $\alpha ,\beta$ be the roots of the equation $\mathrm{x}^{2} - 3\mathrm{x} + \mathrm{r} = 0$ and $\frac{\alpha}{2}, 2\beta$ be the roots of the equation $\mathrm{x}^{2} + 3\mathrm{x} + \mathrm{r} = 0$ . If the roots of the equation $\mathrm{x}^{2} + 6\mathrm{x} = \mathrm{m}$ are $2\alpha +\beta +2\mathrm{r}$ and $\alpha - 2\beta -\frac{\mathrm{r}}{2}$ , then m is equal to :-

Option A:
$-135$
Option B:
$-567$
Option C:
$135$
Option D:
$567$
Correct

Answer: D

Step-by-step solution

$\alpha+\beta=3$ $\frac{\alpha}{2}+2 \beta=-3$ On solving, we get $\alpha=6 ; \beta=-3$ Product of roots $=\alpha \beta=\mathrm{r} \Rightarrow \mathrm{r}=-18$ Now for $\mathrm{x}^{2}+6 \mathrm{x}-\mathrm{m}=0$ Product of roots $=-\mathrm{m}$ $=(2 \alpha+\beta+2 \mathrm{r})\left(\alpha-2 \beta-\frac{\mathrm{r}}{2}\right)$ $\Rightarrow-\mathrm{m}=(-27)(21)$ $\Rightarrow \mathrm{m}=567$

Answer key and solution verified before publishing.

Practise Quadratic Equations

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Theory of Quadratic Equations