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P4 Topic 1: Inequalities
Modulus inequalities 1 backmore
|ax2+bx+c| > dx+e
The graph of y=2x+2 is shown in blue. The graph of y = |x2-1| is also shown. but where |x2-1| > 2x+2 the graph is red, otherwise it is green.

Therefore the solution to |x2-1| > 2x+2 is x<-1 and x>3

Change the values of a, b, c, d or e and solve |ax2+bx+c| > dx+e


Summary

  • In solving |ax2+bx+c| > dx+e, you need to solve |ax2+bx+c| = dx+e first
JavaMath interactive graphs Based on free Java applets from JavaMath