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Question 48
If $$x$$ and $$y$$ are real numbers such that $$x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$$, then the value $$x - 2y$$ is
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
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