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=
![\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}](https://tex.z-dn.net/?f=%20%5Cfrac%7B-b%2B%2F-%20%5Csqrt%7Bb%5E%7B2%7D-4ac%7D%20%7D%7B2a%7D%20)
so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28
x=
![\frac{-(-8)+/- \sqrt{(-8)^{2}-4(1)(28)} }{2(1)}](https://tex.z-dn.net/?f=%20%5Cfrac%7B-%28-8%29%2B%2F-%20%5Csqrt%7B%28-8%29%5E%7B2%7D-4%281%29%2828%29%7D%20%7D%7B2%281%29%7D%20)
x=
![\frac{8+/- \sqrt{-48} }{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B8%2B%2F-%20%5Csqrt%7B-48%7D%20%7D%7B2%7D%20)
x=
![\frac{8+/- 4 \sqrt{-3} }{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B8%2B%2F-%204%20%5Csqrt%7B-3%7D%20%7D%7B2%7D%20)
x=
![4+/- 2 \sqrt{-3}](https://tex.z-dn.net/?f=%204%2B%2F-%202%20%5Csqrt%7B-3%7D%20)
x=
![4+/- 2i \sqrt{3}](https://tex.z-dn.net/?f=%204%2B%2F-%202i%20%5Csqrt%7B3%7D%20)
there are no real numbers that satisfy this
OH MY GOSH I THINK I GET IT! difference as in division. it wants you to i think make for example make 2 the answer to an equation like 5-3=2! 5 and 3 are prime numbers.
Answer:
I believe your answer would be C
To solve for m means to get it alone on a side
so..
![y = mx + b](https://tex.z-dn.net/?f=y%20%3D%20mx%20%2B%20b)
![y - b = mx](https://tex.z-dn.net/?f=y%20-%20b%20%3D%20mx)