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I may add some questions regarding quadratic equations later..

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- Feb 9th 2011, 12:14 AMYuukiQuadratic Equation Question
Attachment 20721

The question is in the image.

Let me know if you can't see it.

I may add some questions regarding quadratic equations later.. - Feb 9th 2011, 12:46 AMemakarov
What does " " mean? Does it mean "

**or**", or "**and**", or something else? Also, should it say ax^2+bx+cx or ax^2+bx+c? - Feb 9th 2011, 12:53 AMYuuki
It means " and ".

Sorry for the confusion.

And it's supposed to say ax^2+bx+c, sorry. - Feb 9th 2011, 01:04 AMemakarov
This information does not guarantee that . For example, if and , then , but .

- Feb 9th 2011, 01:13 AMYuuki
Again, I have to add & apologize.

If and

Then would factoring be ? - Feb 9th 2011, 01:16 AMYuuki
Sorry, I double posted.

Any way I can delete this?? - Feb 9th 2011, 01:47 AMIthaka
- Feb 9th 2011, 02:22 AMemakarov
Please allow me to continue being a skeptic. I am having some mental block.

Why is it that if , are roots of , then ? The converse is obviously true. I mean that we are completely used to this, but since the question is so basic, the proof should be similarly low-level, based on simple arithmetic facts. - Feb 9th 2011, 02:40 AMHallsofIvy
Because the set of all polynomials, p(x), of degree two or less, such that and , forms a one-dimensional subspace of the vector space of all polynomials of degree two or less. And that is, in my opinion, based on very "low level arthmetic facts"- it's just tedious to say without using the "vector space" terminology.

- Feb 9th 2011, 02:59 AMemakarov
Let's consider only polynomials of degree 2 or less.

This says that there exists a polynomials with the following property. For every , iff there exists an such that .

This fact it pretty similar to the original claim that iff . The question is*why*it holds.

The temptation is precisely to resort to more complex facts, but it is not clear that this does not involve circular reasoning. I would like to see a proof from simpler facts. - Feb 9th 2011, 10:41 AMemakarov
Here is one way, inspired by Little Bézout's theorem. Suppose and and are the only roots of . Then, one can check directly that

Indeed, the right-hand side expands to , but since is a root, .

The roots of the left-hand side are and , and the roots of the right-hand side are and . Therefore, . - Feb 10th 2011, 11:36 PMYuuki
Thank you, emakarov, HallsofIvy, and Ithaka/

I have another question.

Quote:

In the above equation, solve when

When or

When or

I don't understand why in both circumstances ( and )

I know you can say that because the problem says "when ", but I'm not satisfied with that answer.

I think I may be content if I do the same problem with a different way of approaching it... - Feb 11th 2011, 12:04 AMProve It
- Feb 11th 2011, 01:49 AMemakarov
- Feb 11th 2011, 02:02 AMIthaka