The question is in the image.
Let me know if you can't see it.

2. 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?

3. It means "$\displaystyle x=\alpha$ and $\displaystyle x=\beta$".
Sorry for the confusion.
And it's supposed to say ax^2+bx+c, sorry.

4. 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 $\displaystyle a(x-\alpha)(x-\beta)=0$.

5. 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 $\displaystyle a(x-\alpha)(x-\beta)$?

6. Sorry, I double posted.
Any way I can delete this??

7. Originally Posted by Yuuki
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 $\displaystyle a(x-\alpha)(x-\beta)$?
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:

$\displaystyle a(x-\alpha)(x-\beta)=0$

8. 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.

9. 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.

10. Let's consider only polynomials of degree 2 or less.

Originally Posted by HallsofIvy
Because the set of all polynomials, p(x), of degree two or less, such that $\displaystyle p(\alpha)= 0$ and $\displaystyle p(\beta)= 0$, form 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.
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.

11. 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, $\displaystyle \beta=-(b/a+\alpha)$.

12. Thank you, emakarov, HallsofIvy, and Ithaka/

I have another question.

$\displaystyle x^2+mx-6m^2=0$
In the above equation, solve $\displaystyle m$ when $\displaystyle x=-3$
I was able to solve the problem;
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...

13. Originally Posted by Ithaka
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:

$\displaystyle a(x-\alpha)(x-\beta)=0$
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".

14. Originally Posted by Yuuki
$\displaystyle m^2+m-6m^2=0$
In the above equation, solve $\displaystyle m$ when $\displaystyle x = -3$
This does not make much sense since there is no x in the equation. The solutions are m = 0 and m = 1/5.

15. Originally Posted by Prove 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".
Yes, thank you very much. I know the difference between the 2 , it was a typing error made at the time I joined the forum and I overlooked it afterwards.

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