Question

$x^2+y^2+2xy=84$ and $x^2+y^2+\sqrt{xy}=14$ Find, $x$ and $y$?

Answer 1

Given equations
$x^2+y^2+2xy=84$ ..........(1)
$(x+y{)}^2=84$
$x+y=\pm \sqrt{84}$
$x+y=9.1651$ or $-9.1651$..........(2) &
$x^2+y^2+\sqrt{xy}=14$ ........(3)
Subtracting (3) from (1), we get
$2xy-\sqrt{xy}=84-14$
$2(\sqrt{xy}{)}^2-\sqrt{xy}-70=0$
Solving for $\sqrt{xy}$ as follows
$\sqrt{xy}=\frac{-(-1)\pm \sqrt{(-1{)}^2-4(2\left)\right(-70)}}{2\left(2\right)}$
$\sqrt{xy}=\frac{1\pm \sqrt{561}}{4}$

$xy={\left(\frac{1\pm \sqrt{561}}{4}\right)}^2$
$xy=38.0857$ or $32.1643$ .........(4)
Hope you can carry out rest calculations by solving (2) & (4)