help

Here is one way:

x = y+4

xy = 2

(y+4)(y)= 2

y^2 + 4y-2 = 0     Quadratic Formula shows    y = .44949      and  -4.44949

                                                        then x =          4.44949     and   -.44949

|x+y| = 4.89898

If x - y = 4 and xy = 2, then find |x + y|.

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{(x+y)^2-(x-y)^2}{2}} &=& \dfrac{x^2+2xy+y^2-x^2+2xy-y^2}{2} \\\\ \dfrac{(x+y)^2-(x-y)^2}{2} &=& \dfrac{4xy}{2} \\\\ (x+y)^2-(x-y)^2 &=& 4xy \quad | \quad x - y = 4,\ xy = 2 \\\\ (x+y)^2-4^2 &=& 4* 2 \\ (x+y)^2-16 &=& 8 \\ (x+y)^2 &=& 8+16 \\ (x+y)^2 &=& 24 \\ x+y &=& \sqrt{24} \\ x+y &=& \sqrt{4*6} \\ \mathbf{ x+y } &=& \mathbf{2\sqrt{6}} \\ \hline \end{array}\)