math
posted by AnnaMarie .
wow. i can't believe i forget this from last year.
5x^2  11x  36 = 0
theres some kind of box trick or somethin to factor that but i forget!

Three possible tricks
1. quadratic equation
2. complete the square
3. factor it 
factoring
(5x+9)(x4) = 0
x = 9/5
or
x = 4 
i'm french so i don't know how do u solve this equation in usa
5x^2  11x  36 = 0
11²4*5*(36)=121+720=841
sqrt(841)=29
solutions are (1129)/(2*5) and (11+29)/(2*5) 
umm... i don't remember how to do any of it. like to factor and get the ( ) ( ) can you tell me how to do that

ooh.. damon how did you get the (5x+9)(x4) = 0 that's what i need to know

complete the square
5 x^2  11 x = 36
x^2  11/5 x = 36/5
x^2 11/5 x + 121/100 = 36/5 + 121/100
(x11/10)^2 = 841/100
x11/10 = +/ 29/10
x = (11+29)/10 = 4
or
x = (1129)/10 = 18/10 = 9/5 
Oh, ok, I guessed'
Like so
5 is only 5 * 1
so there has to be something like
(5x + ;;;) (x  :::)
now the Plus and minus may change and 36 has factors
2*2*3*3
which could be 12 and 3
or 6 and six
or 9 and 4
turns out 9 and 4 work 
Tin tin is using some form of quadratic equation
x = b/2 +/ (1/2) sqrt{ b^2  4 a c}
if in form
a x^2 + b X + c = 0 
x = b/2a +/ (1/2a) sqrt{ b^2  4 a c}
if in form
a x^2 + b X + c = 0 
By the way, the quadratic equation really is completing the square
a x^2 + b x + c = 0
x^2 + (b/a) x = (c/a)
x^2 + (b/a) x + (b/2a)^2 = (c/a) +(b/2a)^2
(x + b/2a)^2 = [c/a + (b/2a)^2 ]
so
x+ b/2a = +/ sqrt (b^2/4a^2  4 a c/4a^2)
or
x = b/2a +/ (1/2a)sqrt(b^24 a c)