Question

What are the zeros of y = 2x^2 + 3x - 1?

(Find out what values of x make y = 0.)

Answer 1

The solutions of this need to be found with the quadratic formula

Answer 2

Can you show me?

Answer 3

Do you know the quadratic formula?

Answer 4

It is
\(\huge \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Answer 5

Yes, I remember that now.

Answer 6

Where the quadratic equation is written in the form
\(\huge ax^2+bx+c\)
Just plug the values in, and see what you get :)

Answer 7

To make it easier, your equation is in the form
\[y=ax^2+bx+c\]
So replace a, b, c with numbers.

Answer 8

put it in quadratic form. y=ax^2 + bx + c

Answer 9

a=2 b=3 c= -1

Answer 10

you need to factor it, and set them equal to zero.

Answer 11

That's the problem. I'm having trouble factoring this equation.

Answer 12

@hiimchristina , it's not factorable.

Answer 13

Yeah, I see that..I'm not sure if it is factorable.

Answer 14

Cam, do what we said, and use the quadratic equation. It's probably the easist way.

Answer 15

Yeah, it's not... the only factors of 1 are 1 & 1. They can't combine to make 3.

Answer 16

btw, if the discriminant b^2-4ac is not a perfect square, then it won't come out to rationally factorable.

Answer 17

OK, I'll do the quadratic equation. :)

Answer 18

Inside the radical, I get 17. It is not a perfect square.

Answer 19

OK, I'm sort of stuck now. :(
What is the next step?

Answer 20

\[2x^2 + 3x - 1=0\]
\[x=\frac{-3\pm\sqrt{3^2-4\times 2\times -1}}{2\times 2}\]
\[=\frac{-3\pm\sqrt{9+8}}{4}=\frac{-3\pm\sqrt{17}}{4}\] there is noting more you can do

Answer 21

So is the answer "No solution"?

Answer 22

Or did I misunderstand something?

Answer 23

But is that what I write as my answer, or is it "No solution"?

Answer 24

that is the solution

Answer 25

The question asks what the zeros are. But if there are none, you'd put "None" as the answer, right?

Answer 26

Either that or there is a misunderstanding...

Answer 27

I know that, hence the parenthesis in my question. :)
I am just trying to clarify that the answer is "None" since the equation isn't even factorable.

Answer 28

...
I'm just going with my answer: "None."
I don't want to waste any more time. (No offense.)

Answer 29

both @Luis_Rivera and i wrote the answer. not sure why you would think there is no answer since we both got the same one

Answer 30

Read the question. Does your answer answer the question? You're going to have to plug that in, see what happens. :(

Answer 31

For the value of x.

Answer 32

:D

Well... Maybe I'll just try asking again later. I have 2 more questions to ask today, I don't want them all taking as long as this one.