Question

If $\frac{x^2+y^2}{x+y}=4$, then all possible values of $(x-y)$ is given by

• A. $\left[-2\sqrt{2},2\sqrt{2}\right]$
• B. $[-4,1]$
• C. $\left[-4,4\right]$
• D. $\left[-2,2\right]$

Answer 1

Answer: (C)

$\left[-4,4\right]$

$\frac{x^2+y^2}{x+y}=4$

$x^2+y^2=4(x+y)$
$x^2-4x+y^2-4y=0$

$x^2-4x+4+y^2-4y+4-8=0$

$(x-2{)}^2(y-2{)}^2=8$

$(x-2{)}^2=4$ and $(y-2{)}^2=4$

$(x-2)=-2$ and $(y-2)=-2$

$\therefore x=0\text{or}4$ and $y=0\text{or}4$

$\therefore$ if $x=0$ and $y=4$

$x-y=-4$
if $x=4$ and $y=0$

$x-y=4$
$(x-y)\in [-4,4]$