Whenever we do inequalities in the past, we always say that 2x - 4 > 0 means 2x > 4 and x > 2. But, have you wondered why this is so?

The thing is, the graph of y = 2x - 4 is actually a linear graph, with a positive gradient. This means that, as x increases, y will always increase, for good. It will follow the same direction and never turn. For this reason, y will be more than 0 when x > 2.

Quadratic graphs, however, do not follow this trend. A typical "smiley" face graph first comes from a high value, before coming down to a "lowest point" and then turning back up again. Unlike the linear graph, in which we are certain of the direction of the line, the quadratic graph faces a different treatment.

The graph of

y= x2 - 2x - 3

is a graph going from a high value, cutting the x axis at -1, before making a turn at a lowest point somewhere there and then turning back up, cutting the x-axis once again at 3 before then going up for good.

So, in short, the graph is

- Positive all the way until x = -1
- Negative as it goes down to a minimum and comes back to hit the x-axis, between -1 and 3
- Positive from x = 3 onwards

As such, we cannot conveniently break up

(x + 1) (x - 3) > 0

into x + 1 > 0, x - 3 > 0,

because the region x > -1 includes the negative region between -1 and 3. This is unlike regular "equality" equations, where

(x + 1) (x - 3) = 0

leads into

x + 1 = 0, x - 3 = 0.

A good understanding of how quadratic equations work helps us understand the region(s) which a curve is positive, and the region(s) which a curve is negative.

Treatment of higher powers of x, such as 3, also takes this idea into account, but this is only seen at the A Levels and beyond.

As another example, we wish to find the range of values of x for which (x - 1) (x - 3) > 0.

Using the techniques which you have learnt in Sec 2, the x-intercepts of the graph y = (x - 1) (x - 3) are x = 1 and x = 3.

The graph will in fact divide itself into three regions, typical of a "smiley" face graph with two intercepts:

1. A positive region before the first intercept
2. A negative region between the two intercepts
3. A positive region after the second intercept

Now, again we cannot break up

(x - 1) (x - 3) > 0

conveniently into

x - 1 > 0, x - 3 > 0

as explained previously. This is a conceptual error amongst many students.

Because of the three regions described above, we will find that the positive regions are dictated by x < 1 and x > 3.

For a similar reason, the negative region for (x - 1) (x - 3) will be the centre region, and the inequality describing this region is

1 < x < 3

because the regions extend from 1 to 3.

Again, we cannot conveniently break the inequality down into separate inequalities x - 1 < 0, x - 3 < 0.

But what happens if the graph is, say, a "sad" face graph?

How will we go about doing this?

First, let's describe a possible path of such graphs. The graph will start from a very negative value and then goes up to cut the x-axis at the first point. Thereafter, the graph attains a maximum point before falling down to hit the x-axis at the second point and then falling down forever.

From here, you have successfully described the regions.

A sketch of such graphs, such as the one tutor Maxyn has included, often helps us visualise these regions better to make an informed statement for the range of values of x.

If you are not comfortable with remembering a sad face graph, you can also attempt to make them "smiley", but with a different signage of the inequality.

Equations such as

- (x + 1) (x - 3) > 0

can be easily converted into

(x + 1) (x - 3) < 0

by multiplying both sides by -1 and changing the inequality sign for the negative multiplier (you have learnt this rule of changing inequality signs, correct?).

Though these are different graphs with different regions, our solutions for the respective range of values of x will not change.