factor 4
4=1 times 4
2 times 2
they don't add to 2
set up equation
x+y=2
xy=4
first equation, subtract x from both sides
y=2-x
subsitute for y
x(2-x)=4
distribute
2x-x^2=4
add x^2
2x=x^2+4
subtract 2x
0=x^2-2x+4
use quadratic formula which is
if you have ax^2+bx+c=0 then
x=\(\frac{ -b+/-\sqrt{b^{2}-4ac} }{2a}\) so
1x^2-2x+4=0
a=1
b=-2
c=4
x=\(\frac{ -(-2)+/-\sqrt{(-2)^{2}-4(1)(4)} }{2(1)}\)
x=\(\frac{ 2+/-\sqrt{4-16} }{2}\)
x=\(\frac{ 2+/-\sqrt{-12} }{2}\)
we have \(\sqrt{-12}\) and that doesn't give a real solution
therefor there are no real solutions
but if you want to solve fully
x=\(\frac{ 2+/-2\sqrt{-3} }{2}\)
i=\(\sqrt{-1}\)
x=\(\frac{ 2+/-2i\sqrt{3} }{2}\)
x=\(1+/-i\sqrt{3}\)
x=\(1-i\sqrt{3}\) or x=\(1+i\sqrt{3}\) (those are the 2 numbers)
