JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 4

Real and imaginary parts. With $z=x+iy$, the equation is $|z|^2+(2+i)z+k(2+3i)=0$:

• Real: $x^2+y^2+2x-y+2k=0$, a circle with centre $\left(-1,\tfrac12\right)$ and $r^2=\tfrac54-2k$.
• Imaginary: $x+2y+3k=0$, a line.

A solution exists when the line meets the circle:
$$\frac{(3k)^2}5\le\frac54-2k\ \Rightarrow\ 36k^2+40k-25\le0.$$

Interval. $\alpha$ and $\beta$ are the roots, so $\alpha+\beta=-\tfrac{40}{36}$ and $9(\alpha+\beta)=-10$.

(A)