JEE Main 2026 (Online) 2nd April Evening ShiftQuestion 1
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 :
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. Both equations have the same product $r$; the two sums fix $\alpha$ and $\beta$. Formulas used. (1) For $x^2-Sx+P=0$: sum of roots $S$, product $P$Step 1 - The sums. By (1), $$\alpha+\beta=3,\qquad\frac\alpha2+2\beta=-3\ \Longrightarrow\ \alpha+4\beta=-6.$$ Subtract: $3\beta=-9$, so $\beta=-3$, $\alpha=6$, and $r=\alpha\beta=-18$. (The second product $\frac\alpha2\cdot2\beta=\alpha\beta$ agrees.) Step 2 - The new roots. $$2\alpha+\beta+2r=12-3-36=-27,\qquad\alpha-2\beta-\frac r2=6+6+9=21.$$ Their sum is $-6$, as it must be for $x^2+6x-m=0$. Step 3 - Find $m$. The product is $-m=(-27)(21)=-567$, so $m=567$. Answer: (D)