Math

[Mateo]

Ye... =( Trying to figure it out anyways, just for lulz.

[Ad Bot]

[Reactor Worker]

K, I know what to do, just to lazy to do it. Might solve it later if no one gets it.

[Mateo]

Orly? I've never been taught it was possible to get 0 or below without the use of negative numbers. Can't wait to see the answer I probably won't understand. :3

[Rughugger]

Man the things people do to get others to do their homework for them.

[DrMobius]

Guys...  Like I said, it's impossible.  A positive a value means that the parabola will open upwards.  A postive b value will cause the slope of the graph to be positve at the y intercept.  A positive c value will shift the graph verticaly so that it is raised up above y = 0 at the y intercept.  So therefore, with only natural numbers, infact even with whole numbers, the zeros will both be at x values <= 0.
@Ladios, if some teacher is giving you this, tell him that it's a trick question because such a thing is impossible with what you described.
@Everyone else, Really think about it.  If you find something that disproves my above point, please tell me.

[UnDiViDeD]

Jeesh.  I could solve this, but it would take a while.

[DrMobius]

Do it then.  It's a loving quadratic function.  Nothing about one that uses integer vales only should take a while.
http://www.members.shaw.ca/ron.blond/QFA.GF.APPLET/index.html
Have fun.

[UnDiViDeD]

Lord Pie, at 16 years old, you haven't really been taught the correct method of solving this.  The problem with your thinking lies in the lack of education thus far in your life.  That and the fact that the answers are not necessarily integer values.

[Otis Da HousKat]

That and the fact that the answers are not necessarily integer values.

If you ignore the fact that he said that they are I guess you're right.

[UnDiViDeD]

A, B, and C are integers, but not the two values of x.  Ax²+Bx+C is an irreducible function, it has no roots.  To solve this, you would use quadratic formula and the integrate.  You would use case 3 of the P.F. Theorem.

[Muffinmix]

Problem

Ax²+Bx+C needs to equal zero

A, B, C and X cannot be zero
A, B, C and X cannot be negative

It is unsolvable because, no matter what values you put in there, unless you round off it will never result in zero.

[laremere]

Rounding doesn't even work, since C has to be an integer, so the lowest you could get this to would be 1.000.....0002

[DrMobius]

Lord Pie, at 16 years old, you haven't really been taught the correct method of solving this.  The problem with your thinking lies in the lack of education thus far in your life.  That and the fact that the answers are not necessarily integer values.

Explain to me how my lack in education is leading me to the wrong conclusion.  You only assume that I've not been taught a correct method.  In fact, you don't have any idea what kind of education I've had.  Perhaps I see something that you don't but then maybe it's the other way around.  I am aware that the 0s are not integer values as they are between 0 and 1.  My point was that in every instance I tried using natural numbers, there were no zeros in the specified bounds.  I don't believe that Ladios wants imaginaries used at all either.  If you want to disprove me so bad, then show me what I have apparently missed.  Show me the integer values to put in for a, b, and c that will get the answer that we're supposed to be looking for.  If you're going to say I'm wrong, next time, at least bother to back your stuff up.

Problem

Ax²+Bx+C needs to equal zero

A, B, C and X cannot be zero
A, B, C and X cannot be negative

It is unsolvable because, no matter what values you put in there, unless you round off it will never result in zero.

Personally, I wouldn't accept a rounding off to give a correct answer.

[Reactor Worker]

Hmm...there appears to a error with the problem.

The equation Ax²+Bx+C=0 is a parabolic curve. The X's you are looking for are the x-intercepts on the graph of that curve.

If A, B and C are all positive, than the curve must not only open up, but also start above the x-axis...therefore it doesn't have any x-intercepts.

[DrMobius]

non that are positive at least.  But that's what I've been loving saying this whole time.