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

Question

3915 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-28T15:01:10

Given equations

$x^2+y^2+2xy=84\ ..........(1)$

$(x+y)^2=84$

$x+y=\pm\sqrt84$

$x+y=9.1651\ \ or \ \ -9.1651 ..........(2)$ &

$x^2+y^2+\sqrt{xy}=14\ ........(3)$

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

$\sqrt{xy}=\frac{-(-1)\pm\sqrt{(-1)^2-4(2)(-70)}}{2(2)}$

$\sqrt{xy}=\frac{1\pm\sqrt{561}}{4}$

$xy=(\frac{1\pm\sqrt{561}}{4})^2$

$xy=38.0857\ \ or \ \ 32.1643\ .........(4)$

Hope you can carry out rest calculations by solving (2) & (4)

x^2+y^2+2xy = 84 and x^2+y^2+sqrt(xy) = 14x2+y2+2xy=84andx2+y2+√xy=14x^2+y^2+2xy = 84 and x^2+y^2+sqrt(xy) = 14 Find, x and yxandyx and y?