Let $\alpha ,\beta$ be the roots of the equation ${x}^{2}-3x+r=0$, and, $\frac{\alpha }{2},2\beta$ be the roots of the equation ${x}^{2}+3x+r=0$. If the roots of the equation ${x}^{2}+6x=m$ are $2\alpha +\beta +2r$ and $\alpha -2\beta -\frac{r}{2}$, then $m$ is equal to: [JEE Main 2026, 2 Apr (Shift 2)]

Question

YESPYQ · Accepted Answer

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

Let \(\alpha ,\beta\) be the roots of the equation \({x}^{2}-3x+r=0\), and, \(\frac{\alpha }{2},2\beta\) be the roots of…

Practice more JEE Maths PYQs