You want
(3p-2p^2)/(4-p^2) - (3+p)/(2-p) < 0
p(3-2p)/(2-p)(2+p) - (3+p)(2+p)/(2-p)(2+p) < 0
[p(3-2p) - (3+p)(2+p)]/(2-p)(2+p) < 0
(3p - 2p^2 - p^2 - 5p - 6)/(2-p)(2+p) < 0
(3p^2 + 2p + 6)/(2-p)(2+p) > 0
The numerator is always positive.
So, we want the region where 4-p^2 is positive.
That is, -2 < p < 2
Just to check, graph the two functions, and you'll see that this is the case.
Show all work
solve the Polynomial inequality and express the solution in set notation.
(3p-2p^2)/(4-p^2)<(3+p)/(2-p)
3 answers
Factor the polynomials by pulling out the GCF
6r^2+12r-15
6r^2+12r-15
Sorry, I posted in the wrong place :(
0 answers