CAT 2023 Slot 1 · Quant · Question 2

Move everything to the left so the equation reads as something equal to zero.

That gives {x}^{2}+{\left( x-2y-1 \right)}^{2}+4y\left( x+y \right)=0.

Expanding the last term, 4y\left( x+y \right)=4xy+4{y}^{2}.

Now {x}^{2}+4xy+4{y}^{2} is a perfect square: it is {\left( x+2y \right)}^{2}.

So the equation becomes {\left( x+2y \right)}^{2}+{\left( x-2y-1 \right)}^{2}=0.

A sum of two squares of real numbers is zero only when each square is zero.

From the second square, x-2y-1=0, so x-2y=1.

That is the second option — and the first square additionally gives x+2y=0, so x=\frac{1}{2} and y=-\frac{1}{4}, which satisfy the original equation.

Shortcut. The question asks only for x-2y, and that expression already appears inside the given equation. Once the rest collapses into a square, the bracket \left( x-2y-1 \right) must vanish on its own — the answer is 1 without ever finding x or y.