AS/A2 Pure 5 Pure 6

P4 Topic 1: Inequalities
Modulus inequalities 2
|ax2+bx+c| > 1/(dx+e)
The graph of y=1/(2x+2) is shown in blue. The graph of y = |x2-1| is also shown. but where |x2-1| > 1/(2x+2) the graph is red, otherwise it is green. Therefore there are three regions where |x2-1| > 1/(2x+2). Change the values of a, b, c, d or e and solve |ax2+bx+c| > dx+e

Summary In solving |ax2+bx+c| > 1/(dx+e), you need to solve |ax2+bx+c| = 1/(dx+e) first Based on free Java applets from JavaMath