JEE Main 2026 — Quadratic Equation Question with Solution

From the first equation $x^2-3x+r=0$, the sum and product of the roots are:

$\alpha +\beta =3$

$\alpha \beta =r$

From the second equation $x^2+3x+r=0$, the sum and product of the roots are:

$\frac{\alpha }{2}+2\beta =-3$

$\left(\frac{\alpha }{2}\right)(2\beta )=r\Rightarrow \alpha \beta =r$

We have a system of linear equations in terms of $\alpha$ and $\beta$:

$\alpha +\beta =3$

$\alpha +4\beta =-6$

Subtracting the first equation from the second gives:

$3\beta =-9\Rightarrow \beta =-3$

Substituting $\beta =-3$ into $\alpha +\beta =3$ gives:

$\alpha =6$

Now, we can find $r$:

$r=\alpha \beta =6\times (-3)=-18$

The roots of the third equation $x^2+6x-m=0$ are given as $2\alpha +\beta +2r$ and $\alpha -2\beta -\frac{r}{2}$. Let us calculate their values:

$R_1=2(6)+(-3)+2(-18)=12-3-36=-27$

$R_2=6-2(-3)-\frac{-18}{2}=6+6+9=21$

The product of the roots of the equation $x^2+6x-m=0$ is $-m$. Therefore:

$-m=R_1{R}_2=(-27)(21)$

$-m=-567\Rightarrow m=567$

Answer: $567$