GATE 2026 EE GA Q.3: solution and answer

Official answer: ASource: Official answer key, GATE 2026 EE (IIT Guwahati)

1. At a point of intersection, both equations hold, so the two expressions for $y$ must be equal: $x^2=-x^2-2x-1$.

2. Move every term to one side: $2x^2+2x+1=0$.

3. A quadratic $a{x}^2+bx+c=0$ has real roots only if its discriminant $D=b^2-4ac\ge 0$. Here $a=2$, $b=2$, $c=1$, so $$D=2^2-4(2)(1)=4-8=-4<0.$$

4. Because $D<0$, no real $x$ satisfies both equations. The curves therefore have no common point in the real $(x,y)$ plane.

5. Check by completing the square: $y=-x^2-2x-1=-(x+1{)}^2\le 0$, and it equals zero only at $x=-1$. Also $y=x^2\ge 0$, and it equals zero only at $x=0$. Any common point would need $y=0$, but the two curves reach $y=0$ at different values of $x$ ($x=-1$ and $x=0$), so they never meet.

6. Number of intersection points $=0$, which is option (A).