PDA

View Full Version : Help factoring

Kaner3339
07-23-2011, 09:45 PM
The question is, X has a uniform distribution on (0,1) find P(X+(1/X)> 3). It gets down to P(X^2 -3X +1 >0) and then this equals P((X- ((3-5^(1/2))/2))*(X-((3+5^)1/2))/2). Can someone point me real quick to a link that teaches this factoring? I'm an algebra noob.

Mr. Micro
07-23-2011, 09:55 PM

TheGillotine
07-23-2011, 09:58 PM
P((X- ((3-5^(1/2))/2))*(X-((3+5^)1/2))/2).
holy god that's painful to look at

Gandalf
07-23-2011, 09:59 PM
You probably know how to do this, just don't recognize the context: this is just the quadratic formula: the zeros of ax^2 + bx + c are (-b + sqrt(b^2-4ac))/2a and (-b - sqrt(b^2-4ac))/2a. Why are those the zeros? Because you can factor ax^2 + bx + c into [x-(-b + sqrt(b^2-4ac))/2a] times [x-(-b - sqrt(b^2-4ac))/2a] times a; and the product is 0 if and only if one of the factors is 0.

Googling "quadratic formula" will tell you how to derive it. They won't really get into factoring it; but they'll explain how to derive the quadratic formula by completing the square.

edit: ninja'd in the basic idea by Mr. Micro, but I wrote a lot more detail

Sinkingfun
07-23-2011, 10:00 PM

{3 (+ -) sqrt( (3)^2 -4)}/2

critical points are:

x = 2.618 or x = .3819

therefore, x^2 -3x + 1 > 0 when x is greater than 2.618 or less than .3819.

However, x is uniform on (0,1) so we can disregard the 2.618 case.

therefore we have:

integral 1 from x = 0 to .3819.

TheGillotine
07-23-2011, 10:02 PM
But yeah, I think that was done using the quadratic formula.