Question

Find the zero's of this quadratic.
x^2=4x+8

Answer 1

\[x ^{2}=4x+8\]

Answer 2

@jim_thompson5910 what do you get for this one?

Answer 3

were you able to get any answers at all?

Answer 4

yes

Answer 5

what did you get

Answer 6

\[2 \pm -2\sqrt{3}\]

Answer 7

what did you get?

Answer 8

nailed it again

Answer 9

x^2=4x+8


x^2-4x=8


x^2-4x-8 = 0


Use the quadratic formula to solve for x


\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]


\[\Large x = \frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-8)}}{2(1)}\]


\[\Large x = \frac{4\pm\sqrt{16-(-32)}}{2}\]


\[\Large x = \frac{4\pm\sqrt{16+32}}{2}\]


\[\Large x = \frac{4\pm\sqrt{48}}{2}\]


\[\Large x = \frac{4+\sqrt{48}}{2} \ \text{or} \ x = \frac{4-\sqrt{48}}{2}\]


\[\Large x = \frac{4+4\sqrt{3}}{2} \ \text{or} \ x = \frac{4-4\sqrt{3}}{2}\]


\[\Large x = 2+2\sqrt{3} \ \text{or} \ x = 2-2\sqrt{3}\]


So I'm getting the same thing.

Answer 10

thanks so much! I did it a little differently but I guess it didn't matter :)

Answer 11

yeah as long as you get the same final answer, that's all that matters