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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,520

But to get that you had to introduce new points.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

x0 x1 y0 y1 are the same intersection points. the other one for each polynomial will be the known point.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,520

Yes, but that polynomial needs 3 known points to be a unique quadratic. You have 2 and one unknown point. You can not determine that quadratic knowing only two.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,596

Hi Herc11

Do you have a test example we could try?

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,520

What is wrong with post #9? Can anyone get A and B?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,596

I need a of four quadratics.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

Ok. You are right. But I am confused with the equation of defining a2.

In equation of a2 only the 4 variables x0 x1 y0 y1 exist.

And

if I have four equations for a2 (I do not know the 4 polynomials, only that they have the same 2 inters. points x0 y0 x1 y1 (unknowns),

I know

each polynomial's leading coef. i.e. a2

and one point of each of the polynomials.

Then from the set of 4 equations of a2, x0..y1 can be defined?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,520

Me too, and putting more points on the drawing does not give them to me.

I need more info about those quadratics!

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,596

I need you to give me the leading coefficient of two more quadratics through those two unknown points and a point on each of those.

*Last edited by anonimnystefy (2013-06-20 01:50:27)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

anonimnystefy wrote:

I need a of four quadratics.

So, if you have 4 quadratics plus a point of each of them you can solve it? i.e you can find x0...y1?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,520

For post #9? Yes any additional info would help. I have been asking the OP to state the full problem. It might be possible to get more info out of it.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,596

Herc11 wrote:

anonimnystefy wrote:

I need a of four quadratics.

So, if you have 4 quadratics plus a point of each of them you can solve it? i.e you can find x0...y1?

I think so.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,520

From just the first coefficient?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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First coefficient of four quadratics and a point on each.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Want to use post #9 and I will provide the two more quadratics since I know what A and B is?

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

anonimnystefy

I think so.

I have the same opinion.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,520

It is more information than you provided in post #1 so I am willing to post the challenge. It is worth a shot.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

bobbym wrote

Want to use post #9 and I will provide the two more quadratics since I know what A and B is?

What do you mean exactly?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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I know the answer for post #9 only because I created it. No one else can know it from that information ( maybe ). I can give 2 more quads.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,596

Hi bobbym

Where does it say that we have only 2 quadratics in the first post??

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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It does not, we just went from there. But so far no one has proved it is possible with n quadratics.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

It does not, we just went from there. But so far no one has proved it is possible with n quadratics.

Thats why I am asking.

It seems to me that If there are 4 quad. you can define the 2 interesection points.

Similarly, if the polynomials were of degree 3, 6 polynomials of 3 degree might be needed...and so on...

But I am not sure

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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I am going to say that with 4 quadratics or 176 of them you still can not define the two points. That is my opinion.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,596

bobbym wrote:

I am going to say that with 4 quadratics or 176 of them you still can not define the two points. That is my opinion.

Okay, then give me the first coefficient of four quadratics and a point on each of them and I will try finding the 2 intersections.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,520

Yes, give some time. I am very interested in your result. Please hold.

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