#x^2+y^2+2xy = 84 and x^2+y^2+sqrt(xy) = 14# Find, #x and y#?

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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)