Solve 1/(2x+1) > x/(3x-2)
I've so far done 1/(2x+1) - x/(3x-2) > 0
(2x^2 + 2x -2)/(2x+1)(3x-2) > 0
Now I know that I have to make the top line zero, but it doesn't factorise. Any help?
Thanks.
Hint: Quadratic Formula.
You wouldn't be studying for an FP2 exam would you by any chance ?
Try multiplying both sides by $\displaystyle (2x+1)^2(3x-2)^2$
This gives you $\displaystyle (2x+1)(3x-2)^2 > x(2x+1)^2(3x-2)$
A little rearranging gives you $\displaystyle (2x+1)(3x-2)((3x-2) - x(2x+1)) > 0$
Giving you critical values of $\displaystyle \frac{-1}{2}$ and $\displaystyle \frac{2}{3}$, if you look at the discriminant of the other equation you will notice that it is less than zero, there no real roots.
Using your graphic calculator if you have one, draw the graph of $\displaystyle y = \frac{1}{2x+1} - \frac{x}{3x-2}$, and look to see when it satisfies the inequality.
Hope this helps