CAT 2023 Slot 1 QA Question & Solution
AlgebraMedium
Question
If $x$ and $y$ are real numbers such that $x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$, then the value $x - 2y$ is
Options
0
1
-1
2
Solution
Given, $x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$
=> $x^2+4xy+4y^2+\left(x-2y-1\right)^2=0$
=> $\left(x+2y\right)^2+\left(x-2y-1\right)^2=0$
For the L.H.S. of the equation to be 0, each of the square terms should be 0 (as squares cannot be negative)
=> x - 2y - 1 = 0 => x - 2y = 1