Examples of Solving Quadratic Inequalities

This inequality has one quadratic being greater than another quadratic. I could try working with the two associated functions, y₁ = 2x² + 4x (shown in green on the graph below) and y₂ = x² − x − 6 (shown below in blue).

Here's the graph:

As you can see, this graph is not particularly helpful. I really can't tell *exactly* where the two lines cross over each other, so intervals are hard to pin down.

On the other hand, if I start properly by isolating the quadratic, my life will be much simpler.

This last inequality is simpler to deal with because now all I have to do is find where x² + 5x + 6 equals zero (which is easy) and then pick the correct intervals based on just the one parabola (which is also easy).

In other words, it is simpler to compare one parabola with the x-axis than it is to *attempt* to compare two parabolas with each other. But the solution to the simpler one-parabola inequality will be the same as the solution to the original two-parabola inequality, so I'll work with the simpler case.

The zeroes (that is, the x-intercepts) of the isolated quadratic are at:

These two intercepts split the number-line into three intervals, namely:

On which of these three intervals is y = x² + 5x + 6 above the x-axis? Since this positive quadratic graphs as a right-side-up parabola, then the quadratic is above the axis on the ends:

So the solution is:

The quadratic is already isolated on the left-hand side of the inequality, so I can go straight to finding the zeroes:

These x-intercepts split the number line into three intervals:

Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x² + 2x − 8 is a right-side-up parabola, the graphed line is below the x-axis on the interval in the middle (that is, on the interval that's between the two intercepts).

Since this is an "or equal to" inequality, the boundary points of the intervals (that is, the x-intercepts themselves) are included in the solution.

So my answer is:

First, I'll isolate the quadratic on one side of the inequality, so I'll have zero on the other side:

Now I want to find the zeroes. There are no factors of +7 that add to −6, so this quadratic doesn't factor. Instead, I'll use the Quadratic Formula:

The "minus" zero is smaller than the "plus" zero; the approximate decimal values of these zeroes (if you were graphing this) are 1.58579 and 4.41421.

This is a positive quadratic, so it's an upward-opening parabola. I'm needing to find the interval(s) over which the quadratic is negative.

But I don't need a graph to find the solution interval(s); I know that this quadratic's upward-opening parabola can be below the x-axis only in the middle of the graph, between the two x-intercepts. So my answer, in interval notation, is: