Question 71

If $$x^{2}$$ + $$y^{2}$$ + 6x + 5 = 4(x - y) then x - y is

Solution

$$x^{2} + y^{2} + 6x + 5 = 4(x - y)$$

=$$x^{2} + y^{2} + 2x + 4y + 5 = 0$$

=$$x^{2} + 2x + 1 + y^{2} + 4y + 4 = 0$$

=$$(x + 1)^2 + (y + 2)^2$$ = 0

So , x= -1 , and y= -2

x - y = -1 + 2 =1

So, the answer would be option a)1.

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