AS/A2 Pure 5 Pure 6

P4 Topic 1: Inequalities
Quadratic inequalities
ax2+bx+c > dx+e
The solutions to inequalities like ax2+bx+c > dx+e can be illustrated graphically. The line y=x+1 is shown in blue. The graph of y = x2-4x+5 is also shown, but where x2-4x+5 > x+1 the curve is coloured red, otherwise it is green. Therefore the solution to x2-4x+5 > x+1 is x<0 and x>4 Make a=-1, b=-2, c=2, d=-1, e=4 and hence deduce the solution to -x2-2x+2 > -x+4 Change the values of a, b, c, d or e and solve ax2+bx+c > dx+e

Summary When a=-1, b=-2, c=2, d=-1, e=4, the quadratic curve is never above the line, so there is no solution. When solving any inequality ax2+bx+c > dx+e, you will first need to solve the equation ax2+bx+c = dx+e Note the difference that positive or negative a makes to the solution Based on free Java applets from JavaMath