Respuesta :

the answer is 125 and -120 gthe thing is 15000 is positive
ok so there are no real solutions for xy=15000 and x+y=5
there are immaginary solutions though so

xy=15000
x+y=5


x+y=5
y=5-x

x(5-x)=15000
-x^2+5x=15000
0=x^2-5x+15000
quadriacic formula
if ax^2+bx+c=0, then
x=[tex] \frac{-b+/- \sqrt{b^{2}-4ac} }{2a} [/tex]

x^2-5x+15000
a=1
b=-5
c=15000
x=[tex] \frac{-(-5)+/- \sqrt{(5)^{2}-4(1)(15000)} }{2(1)} [/tex]
x=[tex] \frac{5+/- \sqrt{25-60000} }{2} [/tex]
x=[tex] \frac{5+/- \sqrt{-59975} }{2} [/tex]
x=[tex] \frac{5+/- 5i\sqrt{2399} }{2} [/tex]
so the numbers are
[tex] \frac{5+ 5i\sqrt{2399} }{2} [/tex] and [tex] \frac{5- 5i\sqrt{2399} }{2} [/tex]




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