By the end of this section, you should be able to:

1. Solve linear inequalities and plot their results on a number line

2. Understand how to solve a quadratic inequality

The cool thing about linear inequalities is that if you have been following along so far, you will already know how they work. For example, when we discussed how a radical function, such as \(\sqrt{x}\) cannot have a negative number inside the radical, we defined its domain as being all numbers **greater than or equal** to zero, as represented by the inequality symbol \(\geq\). In this section, we will begin with linear inequalities then briefly skim over quadratic inequalities.

For example, if we were asked to solve the inequality \(4x - 8 > 0\), how would we go about doing so?

Well, let's start by putting the inequality into words. This inequality is asking us: "for what values of x is \(4x-8\) greater than zero"?

We can solve this algebraically just as we would solve for \(x\), acting as if the \(>\) symbol was replaced by =:

\(4x - 8 > 0\)

\(4x > 8\)

\(x > 2\)

Therefore, 4x - 8 is greater than zero for all values of x greater than two. We can check this by graphing the function:

As we can see from the graph, the function is increasing as x increases and is equal to zero at the point \((2,0)\). Therefore, for all points with x greater than 2, y is greater than zero, so our math is correct. Sometimes, we might be asked to plot this inequality on a number line. For this example, the number line would be:

Note that the point \(x=1\) is not included in the inequality because we are looking only for values greater than; not equal to!

Now, another example: \(-3 \leq x-1 \leq5\).

This inequality is asking us: "for what values of \(x\) is \(x-1\) between -3 **and** 5, inclusive. We add the word "inclusive" when we include the endpoints (5 and -3 in this case).

We can solve this type of inequality in one of two ways. First, we can treat it as two separate inequalities, \(-3 \leq x-1\) and \(x-1 \leq 5\).

Solving as we did before, we get \(-2 \leq x\) and \(x \leq 6\). Combining the two inequalities, we get \(-2 \leq x \leq 6\), or x between -2 and 6.

Another way to solve is by solving for x in both parts of the inequality at the same time. For this example:

\(-3 \leq x -1 \leq 5\)

\(-3 \color{red}{+ 1} \leq x - 1 \color{red}{+ 1} \leq 5 \color{red}{+1}\) adding x to all three parts of the inequality to remove the -1 term from \(x-1\).

\(-2 \leq x \leq 6\)

Plotting this on a number line, we get:

In the previous example, we were asked for a composite of two inequalities joined by "and". In the following case, we will solve one where the inequalities are separated by "or":

Solve the inequality: \(13 -x \geq 5\) or \(13 - x \leq 1\).

In words, this is asking us to find the points for which the line \(13-x\) is greater than *or equal to* 5 **or** less than *or equal to* one. Solving as we did before, we get:

\(13 -x \color{red}{-5} \geq 5 \color{red}{-5}\)

\(8 - x \geq 0\)

\(8 \geq x\) This is one of two solutions

\(13 -x \color{red}{-1} \leq 1 \color{red}{-1}\)

\(12 - x \leq 0\)

\(12 \leq x\) and this is the second!

Therefore, the solution is that x must be less than 8 or greater than 12; \(8 \geq x\) or \(12 \leq x\). Expressing this as a number line, we get the image below. Note that we include the endpoints 8 and 12 in this example.

Finally, quadratic inequalities. At this point, we can solve such inequalities in one of two ways: by reason, or by graphing.

For example, we might be asked to find the range for which \((x-2)(x-3) \geq 0\).

Employing our thinking caps as pictured in the figure below (substitute your face for added effect), we can reason that because there is no negative coefficient before the quadratic function's equation, its concavity is upwards. Therefore, we know that between the points 2 and 3 (the function's zeroes), \(f(x) \lt 0\). Therefore, for x greater than or equal to 3 (the right zero point) and x less than or equal to 2 (the left zero point), the function is greater than or equal to zero. Writing this as an inequality, we get \(x \leq 2\) **or** \(x \geq 3\).

Next, we can verify our prediction by graphing:

We see that we were right! The function is greater than or equal to zero when x is greater than or equal to 3 or less than or equal to 2.

That's it for this lesson. See you next time

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