Math

[Ladios]

Ax²+Bx+C=0.
A B and C are all positive integers and X is two numbers where 0<x<1.
Find A, B, and C where A+B+C= the smalest numbers you can get and both values of x.

Whoever finds out what this is gets a prize.
Possibly money but such a prize would be delayed by at least a couple weeks, but you would get it nonetheless... maybe $10-$20(US)


Also even if you find the answer but I am able to find it on google before tomorrow then no one wins.

[Ad Bot]

[XaMMaX211]

[DrMobius]

x is between 0 and 1 and A, B, and C are just some random integers?
i²x²+x+0?  0 at x=0, x=1  Never said stuff about imaginaries  :D
Do I win?

[Gamefandan]

Ax²+Bx+C=0.
A B and C are all positive integers and X is two intregals where 0<x<1.
Find A, B, and C where A+B+C= the smalest numbers you can get and both values of x.

Whoever finds out what this is gets a prize.
Possibly money but such a prize would be delayed by at least a couple weeks, but you would get it nonetheless... maybe $10-$20(US)


Also even if you find the answer but I am able to find it on google before tomorrow then no one wins.

HA! I'm working on the same kind of math at school.

[Sirrus]

                _______
X=(-B +/-V B^2-4ac )/2a=0

?

[Ladios]

x is between 0 and 1 and A, B, and C are just some random integers?
i²x²+x+0?  0 at x=0, x=1  Never said stuff about imaginaries  :D
Do I win?

Not at all. For one, i and 0 are not positive integers. Two, x is greater than 0 and less than 1. It can't be either.

[Gamefandan]

x = ( -B +- SQRT( B2 - 4AC ) ) / 2A

Ok send me my money in an armored car.

[FlyGuy45]

Let's use the Quadratic formula, with 4 unknowns!

[black knight]

( -B +- SQRT( B2 - 4AC ) ) / 2A

[Muffinmix]

                _______
X=(-B +/-V B^2-4ac )/2a=0

?

Well you'll need a little bit of derivation, but you're close.

We have 3 unknowns, so we need 3 different equations to solve this. We have a range which basically means X<1 and X<0

1>(-B +/-V B^2-4ac )/2a

0<(-B +/-V B^2-4ac )/2a

And the kicker,

Ax2+Bx+C=0

That's all I'm giving you guys for now, I want that money so.

We also have x but I know how to deal with it

[Otis Da HousKat]

X is two intregals where 0<x<1.

You can only take an integral in a closed set. Since a definite integral is the area of a function under the curve.

But, you could replace the open points with the limit at those points(from their respective sides) to give you an answer.

Of course none that would help with the answer.


*fake edit*

loving durrr, he said positive integers. wait misread never mind.

[Gamefandan]

( -B +- SQRT( B2 - 4AC ) ) / 2A

x = ( -B +- SQRT( B2 - 4AC ) ) / 2A

qft

[DrMobius]

I have trouble believing that there is a real answer.  As I stated earlier, it can be done ith imaginaries.  There are no factored forms ((x+a)(x+b)) that I have tryed that can produce only positve coefficients while still producing the positive numbers that you are looking for, let alone any positive numbers.  So my answer is that it is, assuming that I can use only real whole numbers, and have zeros at 0<x<1, is that is not possible.

You can only take an integral in a closed set. Since a definite integral is the area of a function under the curve.

But, you could replace the open points with the limit at those points(from their respective sides) to give you an answer.

Of course none that would help with the answer.


*fake edit*

loving durrr, he said positive integers. wait misread never mind.

I'm assuming that he made a typo.

[Hugums]

42

[Ladios]

I don't really know what a loving intregal is. The point is to find the numbers for x that are between 0 and 1 exclusive.