## Pre-Calculus For Dummies, 2nd Edition (2012)

### Part I. Set It Up, Solve It, Graph It

### Chapter 2. Playing with Real Numbers

*In This Chapter*

Working with equations and inequalities

Mastering radicals and exponents

If you’re taking a pre-calculus class, you’ve already taken Algebra I and II and survived (whew!). You may also be thinking, “I’m sure glad that’s over; now I can move on to some new stuff.” Although pre-calculus presents many new and wonderful ideas and techniques, these new ideas build on the solid-rock foundation of algebra. Alas, we must refresh your memory a bit and test just how sturdy your foundation is.

We assume that you have certain algebra skills down cold, but we begin this book by reviewing some of the tougher ones that become the fundamentals of pre-calculus. In this chapter, we review solving inequalities, absolute-value equations and inequalities, and radicals and rational exponents. We also introduce a new way to express solution sets: interval notation.

**Solving Inequalities**

By now you’re familiar with equations and how to solve them. Pre-calculus teachers generally assume that you know how to solve equations, so most courses begin with inequalities. An *inequality *is a mathematical sentence indicating that two expressions aren’t equal. The following symbols express inequalities:

Less than: <

Less than or equal to: ≤

Greater than: >

Greater than or equal to: ≥

**Recapping inequality how-tos**

Inequalities are set up and solved the same way as equations; the inequality sign doesn’t change the method of solving. In fact, to solve an inequality, you treat it exactly like an equation — with one exception.

If you multiply or divide an inequality by a negative number, you must change the inequality sign to face the opposite way.

For example, if you must solve –4*x* + 1 < 13, your work follows these steps:

–4*x* < 12

*x* > –3

You first subtract 1 from both sides and then divide both sides by –4, at which point the less-than sign changes to the greater-than sign. You can check this solution by picking a number that’s greater than –3 and plugging it into the original equation to make sure you get a true statement. If you check 0, for instance, you get –4(0) + 1 < 13, which is a true statement.

Switching the inequality is a step that many students forget. Look at an inequality with numbers in it, like –2 < 10. This statement is true. If you multiply 3 on both sides, you get –6 < 30, which is still true. But if you multiply –3 on both sides — and don’t fix the sign — you get 6 < –30. This statement is false, and you always want to keep the statements true. The only way for the equation to work is to switch the inequality sign to read 6 > –30. The same rule applies if you divide –2 < 10 by –2 on both sides. The only way for the problem to make sense is to read 1 > –5.

**Solving equations and inequalities when absolute value is involved**

If you think back to Algebra I, you’ll likely remember that an absolute-value equation usually has two possible solutions. Absolute value is a bit trickier to handle when you’re solving inequalities. Similarly, though, inequalities have two possible solutions:

One where the quantity inside the absolute-value bars is greater than a number

One where the quantity inside the absolute-value bars is less than a number

In mathematical terminology, the inequality |*ax* ± *b*| < *c* — where *a, b,* and *c* are real numbers — always becomes two inequalities:

*ax* ± *b* < *c* AND *ax* ± *b* > *–c*

The “AND” comes from the graph of the solution set on a number line, as seen in Figure 2-1a.

The inequality |*ax* ± *b*| > *c* becomes

*ax* ± *b* > *c* OR *ax* ± *b* < *–c*

The “OR” also comes from the graph of the solution set, which you can see in Figure 2-1b.

**Figure 2-1:** The solution to |*ax* ±*b*| < *c* and |*ax* ±*b*| > *c.*

Here are two caveats to remember when dealing with absolute values:

**If the absolute value is less than (<) or less than or equal to () a negative number, it has no solution.** An absolute value must always be zero or positive (the only thing less than negative numbers is other negative numbers). For instance, the absolute-value inequality |2*x* – 1| < –3 doesn’t have a solution, because the inequality is less than a negative number.

Getting 0 as a possible solution is perfectly fine. It’s important to note, though, that having no solutions is a different thing entirely. No solutions means that no number works at all, ever.

**If the result is greater than or equal to a negative number, the solution is all real numbers**. For example, given the equation |*x* – 1| > –5, *x* is all real numbers. The left-hand side of this equation is an absolute value, and an absolute value always represents a positive number. Because positive numbers are always greater than negative numbers, these types of inequalities always have a solution. Any real number that you put into this equation works.

To solve and graph an inequality with an absolute value — for instance, 2|3*x* – 6| < 12 — follow these steps:

**1. Isolate the absolute-value expression.**

In this case, divide by both sides by 2 to get |3*x* – 6| < 6.

**2. Break the inequality in two.**

This process gives you 3*x* – 6 < 6 and 3*x* – 6 > –6. Did you notice how the inequality sign for the second part changed? When you switch from positives to negatives in an inequality, you must change the inequality sign.

Don’t fall prey to the trap of changing the equation inside the absolute-value bars. For example, |3*x* – 6| < 6 doesn’t change to 3*x* + 6 < 6 or 3*x* + 6 > –6.

**3. Solve both inequalities.**

The solutions to this problem are *x* < 4 and *x* > 0.

**4. Graph the solutions.**

Create a number line and show the answers to the inequality. Figure 2-2 shows this solution.

**Figure 2-2:** The solution to 2|3*x*– 6| < 12 on a number line.

**Expressing solutions for inequalities with interval notation**

Now comes the time to venture into interval notation to express where a set of solutions begins and where it ends. *Interval notation* is another way to express the solution set to an inequality, and it’s important because it’s how you express solution sets in calculus. Most pre-calculus books and some pre-calculus teachers now require all sets to be written in interval notation.

The easiest way to find interval notation is to first draw a graph on a number line as a visual representation of what’s going on in the interval.

If the coordinate point of the number isn’t included in the problem (for < or >), the interval is called an *open interval. *You show it on the graph with an open circle at the point and by using parentheses in notation. If the point is included in the solution (≤ or ≥), the interval is called a *closed interval,* which you show on the graph with a filled-in circle at the point and by using square brackets in notation.

For example, the solution set –2 < *x* ≤ 3 is shown in Figure 2-3. ** Note:** You can rewrite this solution set as an

*and*statement:

–2 < *x* AND *x* ≤ 3

In interval notation, you write this solution as (–2, 3].

The bottom line: Both of these inequalities *have* to be true at the same time.

**Figure 2-3:** The graph of –2 < *x*≤ 3 on a number line.

You can also graph *or *statements (also known as *disjoint sets *because the solutions don’t overlap). *Or* statements are two different inequalities where one or the other is true. For example, Figure 2-4 shows the graph of *x* < –4 OR *x* > –2.

**Figure 2-4:** The graph of the *or*statement *x *< –4 OR *x *> –2.

Writing the set for Figure 2-4 in interval notation can be confusing. *x* can belong to two different intervals, but because the intervals don’t overlap, you have to write them separately:

The first interval is *x* < –4. This interval includes all numbers between negative infinity and –4. Because –∞ isn’t a real number, you use an open interval to represent it. So in interval notation, you write this part of the set as (–∞, –4).

The second interval is *x* > –2. This set is all numbers between –2 and positive infinity, so you write it as (–2, ∞).

You describe the whole set as (–∞, –4)(–2, ∞). The symbol in between the two sets is the *union symbol* and means that the solution can belong to either interval.

When you’re solving an absolute-value inequality that’s greater than a number, you write your solutions as *or* statements. Take a look at the following example: |3*x* – 2| > 7. You can rewrite this inequality as 3*x* – 2 > 7 OR 3*x *– 2 < –7. You have two solutions: *x* > 3 or *x* < –5/3.

In interval notation, this solution is .

**Variations on Dividing and Multiplying: Working with Radicals and Exponents**

Radicals and exponents (also known as *roots* and *powers*) are two common — and oftentimes frustrating — elements of basic algebra. And of course they follow you wherever you go in math, just like a cloud of mosquitoes follows a novice camper. The best thing you can do to prepare for calculus is to be ultra-solid on what can and can’t be done when simplifying with exponents and radicals. You’ll want to have this knowledge so that when more challenging math problems come along, the correct answers come along also. This section gives you the solid background you need for those challenging moments.

**Defining and relating radicals and exponents**

Before you dig deeper into your work with radicals and exponents, make sure you remember the facts in the following list about what they are and how they relate to each other:

**A radical is a root of a number.** Radicals are represented by the root sign, . For example, if you take the 2nd root of the number 9 (or the

*square root*), you get 3 because 3 · 3 = 9. If you take the 3rd root (or the

*cube root*) of 27, you get 3 because 3 · 3 · 3 = 27. (In equation form, you write ).

The square root of any number represents the principal root (the fancy term for the *positive root*) of that number. For example, is 4, even though (–4)^{2} gives you 16 as well. is –4 because it’s the opposite of the principal root. When you’re presented with the equation *x*^{2} = 16, you have to state both solutions: *x* = ±4.

Also, you can’t take the square root of a negative number; however, you can take the cube root of a negative number. For example, the cube root of –8 is –2, because (–2)^{3} = –8.

**An exponent represents the power of a number.** If the exponent is a whole number — say, 2 — it means the base is multiplied by itself that many times — two times, in this case. For example, 3

^{2}= 3 · 3 = 9.

Other types of exponents, including negative exponents and fractional exponents, have different meanings and are discussed in the sections that follow.

**Rewriting radicals as exponents (or, creating rational exponents)**

Sometimes a different (yet equivalent) way of expressing radicals makes a solution easier to come by. For instance, when you’re given a problem in radical form, you may have an easier time if you rewrite it by using *rational exponents *— exponents that are fractions. You can rewrite every radical as an exponent by using the following property — the top number in the resulting rational exponent tells you the power, and the bottom number tells you the root you’re taking:

For example, you can rewrite or as .

Fractional exponents are roots and nothing else. For example, 64^{1/3} doesn’t

mean 64^{–3} or . In this example, you find the root shown in the denominator

(the cube root) and then take it to the power in the numerator (the first power). So 64^{1/3} = 4.

The order of these processes really doesn’t matter. You can choose either method:

Cube root the 8 and then square that product

Square the 8 and then cube root that product

Either way, the equation simplifies to 4. Depending on the original expression, though, you may find the problem easier if you take the root first and then take the power, or you may want to take the power first. For example, 64^{3/2} is easier if you write it as (64^{1/2})^{3} = 8^{3} = 512 rather than (64^{3})^{1/2}, because then you’d have to find the square root of 262,144.

Take a look at some steps that illustrate this process. To simplify the

expression , rather than work with the roots, execute the following:

**1. Rewrite the entire expression using rational exponents.**

Now you have all the properties of exponents available to help you to simplify the expression: *x*^{1/2}(*x*^{2/3} – *x*^{4/3}).

**2. Distribute to get rid of the parentheses.**

When you multiply monomials with the same base, you add the exponents.

Hence, the exponent on the first term is

So you get *x*^{7/6} – *x*^{11/6}.

**3. Because the solution is written in exponential form and not in radical form, as the original expression was, rewrite it to match the original expression.**

This gives you .

Typically, your final answer should be in the same format as the original problem; if the original problem is in radical form, your answer should be in radical form. And if the original problem is in exponential form with rational exponents, your solution should be as well.

**Getting a radical out of a denominator: Rationalizing**

Another convention of mathematics is that you don’t leave radicals in the denominator of an expression when you write it in its final form — called *rationalizing the denominator.* This convention makes collecting like terms easy, and your answers will be truly simplified.

A numerator can contain a radical, but the denominator can’t. The final expression may look more complicated in its rational form, but that’s what you have to do sometimes.

This section shows you how to get rid of pesky radicals that may show up in the denominator of a fraction. The focus is on two separate situations: expressions that contain one radical in the denominator and expressions that contain two terms in the denominator, at least one of which is a radical.

**A square root**

Rationalizing expressions with a square root in the denominator is easy. At the end of it all, you’re just getting rid of a square root. Normally, the best way to do that in an equation is to square both sides. For example,

if , then or *x* – 3 = 25.

However, you can’t fall for the trap of rationalizing a fraction by squaring the numerator and the denominator. For instance, squaring the top and bottom of

is *not *equivalent to .

Instead, follow these steps:

**1. Multiply the numerator and the denominator by the same square root.**

Whatever you multiply to the bottom of a fraction, you must multiply to the top; this way, it’s really like you multiplied by one and you didn’t change the fraction. Here’s what it looks like:

**2. Multiply the tops and multiply the bottoms and simplify.**

For this example, you get

**A cube root**

The process for rationalizing a cube root in the denominator is quite similar to that of rationalizing a square root. To get rid of a cube root in the denominator of a fraction, you must cube it. If the denominator is a cube root to the first power, for example, you multiply both the numerator and the denominator by the cube root to the 2nd power to get the cube root to the 3rd power (in the denominator). Raising a cube root to the 3rd power cancels the root — and you’re done!

**A root when the denominator is a binomial**

You must rationalize the denominator of a fraction when it contains a binomial with a radical. For example, look at the following equations:

Getting rid of the radical in these denominators involves using the conjugate of the denominators. A *conjugate* is a binomial formed by taking the opposite of the second term of the original binomial. The conjugate of is . The conjugate of *x* + 2 is *x* – 2; similarly, the conjugate of is .

Multiplying a number by its conjugate is really the FOIL method in disguise. Remember from algebra that FOIL stands for first, outside, inside, and last.

So . The middle two terms always

cancel each other, and the radicals disappear. For this problem, you get *x*^{2} – 2.

Take a look at a typical example involving rationalizing a denominator by using the conjugate. First, simplify this expression:

To rationalize this denominator, you multiply the top and bottom by the conjugate of it, which is . The step-by-step breakdown when you do this multiplication is

Here’s a second example: Suppose you need to simplify the following problem:

Follow these steps:

**1. Multiply by the conjugate.**

The conjugate of is .

**2. Multiply the numerators and denominators.**

FOIL the top and the bottom. (Tricky, we know!) Here’s what we did:

**3. Simplify.**

Both the numerator and denominator simplify first to

which becomes

This expression simplifies even further because the denominator divides into every term in the numerator, which gives you .

Simplify any radical in your final answer — always. For example, to simplify a

square root, find perfect square root factors: . Also, you can add and subtract only radicals that are like terms. This means the number inside the radical and the *index* (which is what tells you whether it’s a square root, a cube root, a fourth root, or whatever) are the same.