Trevor Pythagoras

Quadratic inequalities can often cause problems when trying to decide which range of x values are the solutions. As they are quadratics you usually get two values of x and then have to decide whether it is the values of x between the two bounds that are the solutions or the values of x either side.

For example when solving the inequality:
2x²+x-6>0
we get
(2x-3)(x+2)>0

From this you may be tempted to write x<3/2 and x<-2 however this would be incorrect. The solution must be either -2<3/2, or (x<-2 or x>3/2).
This must be the case for any quadratic with two real roots because all the points between the two roots have one sign and these which are outside the roots have another, as can be clearly seen by considering the general graph.

We can determine which of the two solutions to use by either putting in values to see which work or by drawing a sketch, usually drawing a sketch is the best. For the sketch we only need to consider whether the coefficient of x², in this case 2, is positive or negative so we know which way the graph curves. The y intercept of the graph is irrelevant.

So in our case because 2>=0 the sketch is a “happy” face so the y>0 either side of the roots and therefore our solutions for x are:
x<-2 or x>3/2

By David Woodford