Really simple math time.
1
Name:
Anonymous
2008-04-18 15:00
Ok I'm just doing factorisation of a quardratic expression of the form ax²+bx+c
2
Name:
Anonymous
2008-04-18 15:47
Can't be done.
3
Name:
Anonymous
2008-04-18 17:38
>>2
Thanks
Absolutly no factorisation possible?
4
Name:
Anonymous
2008-04-18 17:48
First one doesn't have any nice factorizations, but you can use the quadratic formula to get the roots and then factor it, naturally. The second one is similar, but slightly different because of the presence of y. Pretend the y isn't there, and get the roots of 3x^2 + 7x + 3. Call these roots a and b. Then 3x^2 + 7xy + 3y^2 = 3(x - ay)(x - by).
5
Name:
Anonymous
2008-04-18 18:57
Just use the quadratic formula for it all....else it's so much time waster guessing.
6
Name:
Anonymous
2008-04-18 21:15
7x²+47x+7
7
Name:
Anonymous
2008-04-19 7:58
>>4
>>5
>>6
Thanks for the help guys.
8
Name:
Anonymous
2008-04-19 22:34
ITT, people doing other people's homework
9
Name:
Anonymous
2008-04-20 1:39
>>8
That's what /sci/ is for!
10
Name:
Anonymous
2008-04-20 9:53
>>8
It isn't my homework. I'm a self studying A level student. So it's obviously difficult when I come across things I can't find an answer for.
11
Name:
Anonymous
2008-04-20 11:49
actually the second one just seems like a (ax+by)² defactorization
12
Name:
Anonymous
2008-04-20 12:02
3x²+7xy+3y²
so
3x² = a²x² -> a² = 3
3y² = b²y² -> b² = 3
7xy = 2abxy -> 2ab = 7
that's an impossible system,
>>11, cause ab would be either 3 or -3, and in both cases, it's impossible for 6 = 7 or -6 = 7.
fail
unless you do it so that 7xy = 6xy + xy and then it'd be
3x² + 7xy + 3y² = 3(x+y)² + xy
13
Name:
Anonymous
2008-04-20 12:04
14
Name:
Anonymous
2008-04-20 12:05
>>13
disregard that, I suck cocks
I meant the other way