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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 "$\displaystyle x=\alpha,\beta$" mean? Does it mean "$\displaystyle x=\alpha$

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

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 $\displaystyle ax^2+bx+c=a(x-\alpha)(x-\beta)$. For example, if $\displaystyle a=b=c=1$ and $\displaystyle x=\alpha=\beta=0$, then $\displaystyle ax^2+bx+c=1$, but $\displaystyle a(x-\alpha)(x-\beta)=0$.

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

If $\displaystyle ax^2+bx+c=0\ x=\alpha$ and $\displaystyle x=\beta$

Then would factoring $\displaystyle ax^2+bx+c=0$ be $\displaystyle a(x-\alpha)(x-\beta)$? - Feb 9th 2011, 01:16 AMYuuki
Sorry, I double posted.

Any way I can delete this?? - Feb 9th 2011, 01:47 AMIthaka
Question should read $\displaystyle x=\alpha$ or $\displaystyle x=\beta$, as x cannot be $\displaystyle \alpha$ and $\displaystyle \beta$ at the same time.

In this case, $\displaystyle \alpha $ and $\displaystyle \beta$ are the roots of the quadratic equation, hence equation can be written:

$\displaystyle a(x-\alpha)(x-\beta)=0$ - 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 $\displaystyle \alpha$, $\displaystyle \beta$ are roots of $\displaystyle ax^2+bx+c$, then $\displaystyle ax^2+bx+c=a(x-\alpha)(x-\beta)$? 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 $\displaystyle p(\alpha)= 0$ and $\displaystyle p(\beta)= 0$, 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 $\displaystyle p_0(x)$ with the following property. For every $\displaystyle p$, $\displaystyle p(\alpha)=\p(\beta)=0$ iff there exists an $\displaystyle a$ such that $\displaystyle p(x)=ap_0(x)$.

This fact it pretty similar to the original claim that $\displaystyle p(\alpha)=p(\beta)=0$ iff $\displaystyle p(x)=a(x-\alpha)(x-\beta)$. 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 $\displaystyle a\ne0$ and $\displaystyle \alpha$ and $\displaystyle \beta$ are the only roots of $\displaystyle ax^2+bx+c$. Then, one can check directly that

$\displaystyle ax^2+bc+c=a(x+(b/a+\alpha))(x-\alpha)$

Indeed, the right-hand side expands to $\displaystyle ax^2+bx-(a\alpha^2+b\alpha)$, but since $\displaystyle \alpha$ is a root, $\displaystyle -(a\alpha^2+b\alpha)=c$.

The roots of the left-hand side are $\displaystyle \alpha$ and $\displaystyle \beta$, and the roots of the right-hand side are $\displaystyle \alpha$ and $\displaystyle -(b/a+\alpha)$. Therefore, $\displaystyle \beta=-(b/a+\alpha)$. - Feb 10th 2011, 11:36 PMYuuki
Thank you, emakarov, HallsofIvy, and Ithaka/

I have another question.

Quote:

$\displaystyle x^2+mx-6m^2=0$

In the above equation, solve $\displaystyle m$ when $\displaystyle x=-3$

When $\displaystyle m=1, x=2$ or $\displaystyle x=-3$

When $\displaystyle m=-\frac{3}{2}, x=\frac{9}{2}$ or $\displaystyle x=-3$

I don't understand why $\displaystyle x=-3$ in both circumstances ($\displaystyle m=1$ and $\displaystyle =-\frac{3}{2}$)

I know you can say that because the problem says "when $\displaystyle x=-3$", 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
Completely off topic, but Ithica, you should use $\displaystyle \displaystyle \pi$ in your siggie instead of $\displaystyle \displaystyle \Pi$, because $\displaystyle \displaystyle \pi$ is the ratio of a circle's circumference to its diameter, while $\displaystyle \displaystyle \Pi$ is used as the symbol for "product".

- Feb 11th 2011, 01:49 AMemakarov
- Feb 11th 2011, 02:02 AMIthaka