Quadratic inequalities are not taught in the E Maths syllabus. If this really appears in E Maths, then something is not right.

You would need to understand the graph of quadratic functions to understand this.

4 < 5x² + 8x
0 < 5x² + 8x - 4
5x² + 8x - 4 > 0
(x + 2) (5x - 2) > 0

Now, if we were to draw the graph of y = (x + 2) (5x - 2), how will it look like? Have you learnt to sketch such graphs?

Unlike linear inequalities or regular quadratic equations, we cannot straight away conclude that x + 2 > 0 or 5x - 2 > 0. This is a common mistake amongst students; we cannot draw parallels from ideas associated with linear inequalities and regular quadratic functions.

Drawing the graph of y = (x + 2) (5x - 2) is straightforward. This is a "smiley" face quadratic curve with x-intercepts of -2 and 2/5. The y-intercept is not relevant here.

Solving the inequality (x + 2) (5x - 2) > 0 is akin to finding the region where the graph of y = (x + 2) (5x - 2) lies above the line the x-axis (i.e. we are solving for y > 0).

You will notice that to the left of the x-intercept -2 or to the right of the x-intercept 2/5, the graph exists above the x-axis. We need these.

Our answer would be x < -2 or x > 2/5.

The reason why linear inequalities work the way they are is largely similar.

Suppose we solve x - 3 > 0. Why is it quite obvious that x > 3 will be our solution?

This is because when we draw the graph pf y = x - 3, the line will extend from x = 3 diagonally upwards forever. There is no end to this. Our solution to this would be x > 3.

Numerically, it is also logical. Just input any number greater than 3...of course x - 3 would be positive in that case!

But (x + 2) (5x - 2) is different. Try putting values of x within these regions and see whether they are positive or negative.

• x < -2
• -2 < x < 2/5 (i.e. values of x between -2 and 2/5)
• x > 2/5