Recent content by Leo Consoli

Thank you very much for the help.

That would mean there exist two numbers (m and n) such that m×n=AC m+n=B But AC is odd, and B is also odd, and there isn't a pair of numbers whose sum and product is odd, so its not possible to factorize it, meaning the polynomial has no rational roots. Is this it?

In this case, how would I find parcels of AC that add up to B?

But can't you make more than one second degree polynomial with the values of a, b and c?

I think that's it.

Thank you, I will edit it.

x^2 - x -3 + 2c = 2x(ax+b) x^2 -2ax^2 - 2bx - x - 3 + 2c = 0 x^2(1-2a) -x(1+2b) -3 + 2c =0 Using girard r1+r2 = (1+ 2b)/(1-2a) r1xr2 = (-3 +2c)/(1-2a) After this I am stuck. Thank you.

Yes, there is everything, except for AB.

Anyone has any ideas for solving this? Thanks.

I am not the OP but arent these propabilities all the same?

Hi, thanks, using triangle similarity between OMN and NBK and then with OMN and BPM I found the following values for BN and BP: BN=(b-a×cos^2(alfa))/sin(alfa) BP=cotg(alfa)×(a-b) With these values of the 4 sides of the quadrylateral and of one of the diagonals I think I can find the other...

Hi, sorry for taking so long to answer I have been very busy. Here are the calculations I did, other things I tried led me to these same results so I couldn't proceed. In the triangle OPK: Sin (alpha)=b/PK ----> PK=b/sin(alpha) Cos(alpha)=b/OK ---> OK= b/cos(alpha) In the triangle OMN...

I have to find OA, sorry, I don't have acess to a computer right now, so its hard to post equations, and its late right now and I have to go to sleep, tomorrow I will try to post them.

Homework Statement The angle alfa has its vertex at a point O, from one of its sides the point M is taken from which the perpendicular to the other side is made with the point N. In the same way, from the other side point K is taken and from there the perpendicular to the other side is traced...

Thank you I will edit it. I see, I didnt notice this method. Thank you, if I had thought like this it would have been much easier.