You are not logged in.
Pages: 1
|x+2/x+1|less than or equal to |x-3|
how do i solve this??
I couldn't find any way!
My lecturer did give me a hint saying that there are four combinations of + and -ve signs.
Offline
benleong2008;
I think you can remove the bar as it called in 4 ways. Giving you 4 rational inequalities to solve.
Last edited by bobbym (2009-09-12 10:03:25)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
I have actually tried using that method a long time ago. However, it always come up with a trinomial at the numerator which is not factorable.
x+2 x+2
----- ≤ x-3 ------ ≤ -x+3
x+1 x+1
x+2 x+2
----- -x+3≤0 ------ +x-3 ≤0
x+1 x+1
x+2-x²-x+3x+3 x+2+x²+x-3x-3
------------------- ≤0 -------------------- ≤0
x+1 x+1
-x²+3x+5 x²-x-1
------------ ≤0 -------- ≤0
x+1 x+1
And when -(x+2/x+1) ≤ x-3 and when -(x+2/x+1) ≤ -x+3, it comes up with -x²+x+1 and x²-3x-5 at the numerator respectively.
Offline
Hi benleong2008;
Try this to get a feel for them:
http://tutorial.math.lamar.edu/Classes/ … ities.aspx
http://www.youtube.com/watch?v=SJecFvUbJOY
http://www.5min.com/Video/Solving-Ratio … -160953981
Of course the trinomial is factorable, that quadratic has roots doesn't it. Just because you can't factor it by eye doesn't mean it can't be factored.
Last edited by bobbym (2009-09-12 16:17:50)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Pages: 1