Question

what are the zeros of the graph y=2x^2 -x -15?

Answer 1

Have you heard of the quadratic formula?

Answer 2

yes

Answer 3

So you have an equation

2x² - x - 15

let your coefficient correspond to the different letters

a = your first coefficient = 2
b = your second coefficient = -1
c = your third coefficient = -15

plug them into that formula

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

This will tell you the zeros (x-intercepts) of the graph

Answer 4

sense its already -1 would it change to poitive1?

Answer 5

I assume you're talking about the -b part? then yes you would

Answer 6

ya so the answer would be 2.81 & -2.56?

Answer 7

\[\frac{ 1 \pm \sqrt{-1^2 - 4(2)(-15)} }{ 2(2) }\]
\[\frac{ 1 \pm \sqrt{1 + 120 } }{ 4 }\]
\[\frac{ 1 \pm \sqrt{121} }{ 4 }\]
\[\frac{ 1 \pm 11 }{ 4 }\]

so breaking that into the 2 equations

\[\frac{ 1 + 11 }{ 4 } =?\]
and
\[\frac{ 1 - 11 }{ 4 } = ?\]

Answer 8

how did you get 121? I redid it and got 119

Answer 9

remember...-1 TIMES -1 = 1 right?

you have -1 times -1 = 1

-4 times 2 times -15 = 120

1 + 120 = 121

Answer 10

so its not just -1^2-4(2)(-15)?

Answer 11

late response

\[\sqrt{-1^2 - 4(2)(-15)}\]

\[\sqrt{1 - (-120)}\]

\[\sqrt{121} = 11\]

Answer 12

I didn't learn it that way

Answer 13

Well that was just 1 portion of the quadratic equation

\[\frac{ 1 \pm \sqrt{-1^2 - 4(2)(-15)} }{ 2(2) }\]

how did you learn it?