Question

Given x^2 + y^2 = 36 and xy = -10 find x + y.

Answer 1

This looks fun

ok so
we will use subsitution
xy=-10
divide both sides by x or y (I will choose y)
x=-10/y

sub -10/y fo x

(-10/y)^2+y^2=36
100/(y^2)+y^2=36
times both sides by y^2
100+y^4=36y^2
minus 36y^2 from both sides
y^4-36y^2+100=0
(y^2)^2-36(y^2)+100=0
quadratic formula
for
ax^2+bx+c=0
x=$\frac{-b+/- \sqrt{b^2-4ac} }{2a}$
x=$\frac{-(-36)+/- \sqrt{(-36)^2-4(1)(100)} }{2(1)}$
x=$18+/- 4\sqrt{14}$

sub
y=-10/x
y=$\frac{-10}{18+/- 4\sqrt{14}}$

so x+y=$18+/- 4\sqrt{14}$+$\frac{-10}{18+/- 4\sqrt{14}}$
multiply first number by $\frac{18+/- 4\sqrt{14}}{18+/- 4\sqrt{14}}$ and add them

x+y=$\frac{(18+/- 4\sqrt{14})(18+/- 4\sqrt{14})-10}{18+/- 4\sqrt{14}}$
or
x+y=$\frac{81+22 \sqrt{14} }{5}$ or $\frac{81-22 \sqrt{14} }{5}$