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
