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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.
|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.
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