Quadratic Equation Question

**Yuuki** · February 9th 2011, 12:14 AM

The question is in the image.
Let me know if you can't see it.
I may add some questions regarding quadratic equations later..

**emakarov** · February 9th 2011, 12:46 AM

What does " $x=\alpha,\beta$ " mean? Does it mean " $x=\alpha$ or $x=\beta$ ", or " $x=\alpha$ and $x=\beta$ ", or something else? Also, should it say ax^2+bx+cx or ax^2+bx+c?

**Yuuki** · February 9th 2011, 12:53 AM

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

**emakarov** · February 9th 2011, 01:04 AM

This information does not guarantee that $ax^2+bx+c=a(x-\alpha)(x-\beta)$ . For example, if $a=b=c=1$ and $x=\alpha=\beta=0$ , then $ax^2+bx+c=1$ , but $a(x-\alpha)(x-\beta)=0$ .

**Yuuki** · February 9th 2011, 01:13 AM

Again, I have to add & apologize.
If $ax^2+bx+c=0\ x=\alpha$ and $x=\beta$
Then would factoring $ax^2+bx+c=0$ be $a(x-\alpha)(x-\beta)$ ?

**Yuuki** · February 9th 2011, 01:16 AM

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

**Ithaka** · February 9th 2011, 01:47 AM

Originally Posted by Yuuki

Again, I have to add & apologize.
If $ax^2+bx+c=0\ x=\alpha$ and $x=\beta$
Then would factoring $ax^2+bx+c=0$ be $a(x-\alpha)(x-\beta)$ ?

Question should read $x=\alpha$ or $x=\beta$ , as x cannot be $\alpha$ and $\beta$ at the same time.

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

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

**emakarov** · February 9th 2011, 02:22 AM

Please allow me to continue being a skeptic. I am having some mental block.

Why is it that if $\alpha$ , $\beta$ are roots of $ax^2+bx+c$ , then $ax^2+bx+c=a(x-\alpha)(x-\beta)$ ? 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.

**HallsofIvy** · February 9th 2011, 02:40 AM

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

**emakarov** · February 9th 2011, 02:59 AM

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 $p(\alpha)= 0$ and $p(\beta)= 0$ , 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 $p_0(x)$ with the following property. For every $p$ , $p(\alpha)=\p(\beta)=0$ iff there exists an $a$ such that $p(x)=ap_0(x)$ .

This fact it pretty similar to the original claim that $p(\alpha)=p(\beta)=0$ iff $p(x)=a(x-\alpha)(x-\beta)$ . 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.

**emakarov** · February 9th 2011, 10:41 AM

Here is one way, inspired by Little Bézout's theorem. Suppose $a\ne0$ and $\alpha$ and $\beta$ are the only roots of $ax^2+bx+c$ . Then, one can check directly that

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

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

The roots of the left-hand side are $\alpha$ and $\beta$ , and the roots of the right-hand side are $\alpha$ and $-(b/a+\alpha)$ . Therefore, $\beta=-(b/a+\alpha)$ .

**Yuuki** · February 10th 2011, 11:36 PM

Thank you, emakarov, HallsofIvy, and Ithaka/

I have another question.

$x^2+mx-6m^2=0$
In the above equation, solve $m$ when $x=-3$

I was able to solve the problem;
When $m=1, x=2$ or $x=-3$
When $m=-\frac{3}{2}, x=\frac{9}{2}$ or $x=-3$

I don't understand why $x=-3$ in both circumstances ( $m=1$ and $=-\frac{3}{2}$ )
I know you can say that because the problem says "when $x=-3$ ", 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...

**Prove It** · February 11th 2011, 12:04 AM

Originally Posted by Ithaka

Question should read $x=\alpha$ or $x=\beta$ , as x cannot be $\alpha$ and $\beta$ at the same time.

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

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

Completely off topic, but Ithica, you should use $\displaystyle \pi$ in your siggie instead of $\displaystyle \Pi$ , because $\displaystyle \pi$ is the ratio of a circle's circumference to its diameter, while $\displaystyle \Pi$ is used as the symbol for "product".

**emakarov** · February 11th 2011, 01:49 AM

Originally Posted by Yuuki

$m^2+m-6m^2=0$
In the above equation, solve $m$ when $x = -3$

This does not make much sense since there is no x in the equation. The solutions are m = 0 and m = 1/5.

**Ithaka** · February 11th 2011, 02:02 AM

Originally Posted by Prove It

Completely off topic, but Ithica, you should use $\displaystyle \pi$ in your siggie instead of $\displaystyle \Pi$ , because $\displaystyle \pi$ is the ratio of a circle's circumference to its diameter, while $\displaystyle \Pi$ 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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