find numbers that multiply to 28 and add them to see if they add to 8
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula
x+y=8
xy=28
x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0
if you have
ax^2+bx+c=0 then x=[tex] \frac{-b+/- \sqrt{b^{2}-4ac} }{2a} [/tex]
so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28
x=[tex] \frac{-(-8)+/- \sqrt{(-8)^{2}-4(1)(28)} }{2(1)} [/tex]
x=[tex] \frac{8+/- \sqrt{-48} }{2} [/tex]
x=[tex] \frac{8+/- 4 \sqrt{-3} }{2} [/tex]
x=[tex] 4+/- 2 \sqrt{-3} [/tex]
x=[tex] 4+/- 2i \sqrt{3} [/tex]
there are no real numbers that satisfy this
