Discriminant

**Fails_at_Math** · August 16th 2009, 11:51 AM

Hi MHF. My question is..

Discuss the number and type of zeroes/roots for each function below by considering the discriminant. [Do not solve]
$p(x)=2x^2+4x+2$
$q(x)=3x^2-6x+2$

I don't really understand what I'm trying to answer here. It would seem I have to calculate the discriminant first. How would I go about doing that? And how does that help me to answer the question?

Thanks.

**Pim** · August 16th 2009, 11:58 AM

You indeed need to calculate the discriminant.

What this question means is actually, how many intersections are there with the x-axis.

You probably are familiar with the fact that if D > 0 there are two intersections, if D = 0, there is one, and if D < 0 there are none. That is it for the number of roots, for the type, I have no idea what they mean either.

(It occurred to me that you might actually not know what they mean with the discriminant, if you don't, please post that and I'll try to help you further.)

**Fails_at_Math** · August 16th 2009, 12:07 PM

Originally Posted by Pim

You indeed need to calculate the discriminant.

What this question means is actually, how many intersections are there with the x-axis.

You probably are familiar with the fact that if D > 0 there are two intersections, if D = 0, there is one, and if D < 0 there are none. That is it for the number of roots, for the type, I have no idea what they mean either.

(It occurred to me that you might actually not know what they mean with the discriminant, if you don't, please post that and I'll try to help you further.)

Mmm, I probably should, but I'm not sure that I do. Something like $b^2-4ac$ ?

**skeeter** · August 16th 2009, 12:09 PM

Originally Posted by Fails_at_Math

Hi MHF. My question is..

Discuss the number and type of zeroes/roots for each function below by considering the discriminant. [Do not solve]
$p(x)=2x^2+4x+2$
$q(x)=3x^2-6x+2$

I don't really understand what I'm trying to answer here. It would seem I have to calculate the discriminant first. How would I go about doing that? And how does that help me to answer the question?

Thanks.

for $f(x) = ax^2+bx+c$ , the discriminant is the value $b^2-4ac$ .

if $b^2-4ac > 0$ ... the function has two real, distinct roots

if $b^2-4ac = 0$ ... the function has one real root of multiplicity two

if $b^2-4ac < 0$ ... the function has two imaginary roots

**VonNemo19** · August 16th 2009, 12:09 PM

Originally Posted by Fails_at_Math

Mmm, I probably should, but I'm not sure that I do. Something like $b^2-4ac$ ?

Yes. That is what the last guy meant by D.

**Pim** · August 16th 2009, 12:13 PM

Okay.

The solutions to intersections with the x-axis of a function in the form:

$<br /> y = ax^2 + bx + c<br />$

are

$x = \frac{-b-\sqrt{D}}{2a}$

and

$<br /> x = \frac{-b+\sqrt{D}}{2a}<br />$

With $D = b^2 - 4ac$

Now, because of the root in this formula, D can tell you about how many solutions there are to the equation. If D is positive, there will be two different answers. If D is zero, it will yield one answer. (As + and - 0 are the same thing) If D is smaller than zero, the root does not exist and there are no intersections with the x-axis.

**Fails_at_Math** · August 16th 2009, 12:20 PM

Ah, ok! Starting to recall this from the lecture. I think I can figure it out now, thank you.

**VonNemo19** · August 16th 2009, 12:26 PM

Hey Pim, I'm gonna jump in here...

Originally Posted by Pim

Now, because of the root in this formula, D can tell you about how many solutions there are to the equation. If D is positive, there will be two different answers.

Because $\pm\sqrt{D}$ will yeild two answers. A positive one, and a negative one.

Originally Posted by Pim

If D is zero, it will yield one answer.

Because zero is neither positive nor negative. And since $\pm0$ is meaningless, you'll only get one answer.

Originally Posted by Pim

If D is smaller than zero, the root does not exist and there are no intersections with the x-axis.

Because you can't take the square root of a negative number.

So, think about it...

**Fails_at_Math** · August 16th 2009, 12:41 PM

So, for $p(x)$ I got $d=0$ , meaning the function crosses the x axis once and has one real root.

$q(x)$ I got $d=-60$ , so the function does not cross the x axis, it has two imaginary roots (or two complex zeroes as I believe he wants it answered).

Is that all the question is asking?

Edit: So if I wanted to actually find the solution to p(x) I could plug it into the quadratic equation which would net me -1 for both + and -. But it seems it doesn't ask for that answer. Correct?

Edit: Made a mistake on q(x), the discriminant is greater than 0.

**VonNemo19** · August 16th 2009, 01:28 PM

Originally Posted by Fails_at_Math

So, for $p(x)$ I got $d=0$ , meaning the function crosses the x axis once and has one real root.

Right on!

Originally Posted by Fails_at_Math

$q(x)$ I got $d=-60$

I'd check that again... $(-6)^2-4(3)(2)=?$

Originally Posted by Fails_at_Math

Is that all the question is asking?

Yes indeed!

Originally Posted by Fails_at_Math

Edit: So if I wanted to actually find the solution to p(x) I could plug it into the quadratic equation which would net me -1 for both + and -. But it seems it doesn't ask for that answer. Correct?

Good stuff!

Originally Posted by Fails_at_Math

Edit: Made a mistake on q(x), the discriminant is greater than 0.

Ok, so you did check it again. Rock on, Yo!

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