Question

Find the x-intercepts:
g(x) = -x^2 + 2x + 3

My book tells me (-1, 0) and (3, 0), but I have no idea how this happened.

The problem before this I solved by factoring. I'm having a brain-fart here and can't seem to be able to get past that negative...
I also tried using the quadratic formula, but I got a complex. Somehow.

...I have no idea why this is so difficult for me right now- I had it down perfectly not even last week!

Answer 1

\[g(x) = -x^2 + 2x + 3\]
The quadratic formula works (if you dont forget the 4)
\[x=\frac{-b±\sqrt{b^2-4ac}}{2a}\]\[x=\frac{-(2)±\sqrt{(2)^2-4(-1)(3)}}{2(-1)}\]\[x=\frac{-2±\sqrt{16}}{-2}\]\[x=1±\frac{4}{-2}\]\[x=1\mp 2\]\[ x_{1,2}=-1,3\]

Answer 2

hence the zeros of \[g(x) \text{ are } (-1,0), (3,0)\]

Answer 3

You're right! I still can't figure out how I got what I did... Probably did forget the 4. Heh.

Thank you!

Answer 4

yeah i tried this and forgot about the four , which kept giving the wrong answer,

Answer 5

Also you can do it by factoring. Product is -3, sum is 2.
(x+1)(-x+3)