High School Algebra II Unlocked (2016)

GOALS

By the end of this chapter, you will be able to

•Solve single-variable rational and radical equations and inequalities

•Write and solve rational and radical equations, inequalities, and functions to describe real-life situations and solve problems

•Graph rational and radical functions

•Use systems of equations to solve single-variable rational and radical equations

Lesson 3.1. Rational Equations in One Variable

REVIEW

Any fraction of the form a/0, where a is any real number, is undefined.

When a factored polynomial is set equal to 0, a variable value that makes one factor equal to 0 makes the entire polynomial equal to 0.

The Rational Root Theorem states that if a polynomial has any rational roots, they must be in the form ±.

The Remainder Theorem states that when a polynomial p(x) is divided by (x − a), the remainder is p(a), so p(a) is equal to 0 if and only if (x − a) is a factor of p(x).

The mean of a set of data is found by adding the values of all elements in the set and dividing that total by the number of elements in the set.

A rational expression is an algebraic expression in the form of a ratio of two polynomials.

A rational equation is an equation that contains one or more rational expressions. These equations can be used to solve problems that involve ratio comparisons or division, such as the calculation of an average.

Will scored a 90 on the math test he took today, which brought his average (arithmetic mean) test grade for the semester up from 78 to 80. How many math tests has he taken this semester, including today’s?

Calculating a mean involves division, so we’ll write a rational equation to represent this situation. Will’s mean test score for the semester is the total value of all of his test scores divided by the number of tests he has taken.

Let n = the number of math tests Will took before today’s test. His average for those first n tests was 78, so 78 = . If you rewrite this equation (multiply both sides by n), you can see that the total of Will’s scores on the first n tests was 78n.

Let’s write an equation for the average of all of Will’s test scores, including today’s. He took a total of n + 1 tests. The total of all of his test scores is the sum of the total of his first n test scores (78n) and his test score from today (90). His new average test score is 80.

80 =

Solve the equation for n.

This isn’t yet our answer, though! Remember, n is the number of tests Will took before today’s test, but the question asks for the total number of tests including today’s. Will has taken a total of 5 + 1 = 6 math tests this semester.

If a rational equation shows one rational expression set equal to another, you can cross-multiply and solve the resulting polynomial equation.

Here’s your chance to
use all the polynomial-
solving techniques you
learned in Chapter 1!

Solve for any real solutions to the following equation.

Cross-multiply and move all terms to one side.

(2x + 1)(x² − 4) = 1(x² − 8x + 50)

2x³ + x² − 8x − 4 = x² − 8x + 50

2x³ − 54 = 0

The shortcut here is to recognize that this is the simple cubic equation 2x³ = 54, or x³ = 27, which has the solution x = 3. If we instead factor 2x³ − 54, we can solve for all three solutions.

The solution to x − 3 = 0 is x = 3. Solving x² + 3x + 9 = 0 requires the quadratic formula.

These two solutions include imaginary numbers, so the only real solution is 3. Test it in the original equation.

The equation is true, so x = 3 is the one real solution to .

Testing your answer in the original equation is an especially good idea when it comes to rational equations, because of the risk of false solutions.

An x-value that makes the denominator of a rational expression equal to zero makes the entire rational expression undefined and therefore cannot be a solution to the given rational equation. However, such x-values may be solutions to the polynomial equations produced during the solution process, through cross-multiplication or other methods. These false solutions are called extraneous solutions.

You’ll specifically want to keep this in mind when dealing with multiple-choice problems like the ones found on the SAT, as extraneous solutions are often offered as trap answers.

Sometimes a rational equation will have multiple potential solutions, including both extraneous solutions and true solutions. You will need to determine which are which by testing each potential solution in the original equation.

Solve:

The least common denominator for all three rational expressions is x(x + 6). Convert all three expressions to have this denominator.

In Lesson 2.4, we
practiced operations on
rational expressions.
Here, we’re using
those skills to combine
rational expressions
in order to solve a
rational equation.

Here is how you may see rational equations on the SAT.

Given the equation above, what is the value of r?

A) 3

B) 6

C) 7

D) There is no such value of r.

The two rational expressions have the same denominator, so their numerators must be equal.

Test each of the potential solutions in the original equation.

When x = 0:

Setting the numerators
equal is the same as
multiplying both sides
by x(x + 6). You could
also have multiplied
every term on both
sides by x(x + 6) at
the beginning of the
solution process.

Two of the rational expressions have 0 in the denominator, so these are undefined. The x-value of 0 is an extraneous solution. It is not a solution to the rational equation.

When x =

This equation is true, so −3 is a solution for x. The only solution to is x = −3.

Solve the following equation for x.

Let’s cross-multiply then solve for possible solutions for x.

The Rational Root
Theorem and the
Remainder Theorem
are what we used to
find the roots of a
cubic polynomial in
Chapter 1, Example 18.
Here, we are using
the same methods
to solve for zeros
of a polynomial
equation, which in
turn are potential
solutions to the given
rational equation.

We need to factor this cubic to solve for all possible values of x. According to the Rational Root Theorem, any rational roots of this cubic will be in the set ±{1, 2, 4, 8}. Let’s use the Remainder Theorem to test some values in the polynomial x³ + 2x² − 4x − 8.

When x = 1, 1³ + 2(1²) − 4(1) − 8 = −9. This is not equal to 0, so 1 is not a root of the polynomial.

When x = 2, 2³ + 2(2²) − 4(2) − 8 = 0. This means that 2 is a root, so (x − 2) is a factor of the cubic. We can use long division to determine the other factor.

We can start by combining the two rational expressions on the left side of the equation. To do so, we must convert them to have their least common denominator, (r − 3)(r + 5), then add them together.

We don’t need to cross-multiply, because the two rational expressions that are equal to one another have the same denominator. This means that their numerators are equal.

r + 5 + r − 3 = 8

2r + 2 = 8

2r = 6

r = 3

It looks like we found the solution!…Or is it an extraneous solution masquerading as a solution? Before you choose (A), test r = 3 in the original equation.

Two of the fractions have denominators of 0, which means that they have undefined values. So, 3 is not a solution to the original rational equation. It is an extraneous solution. There are no other possible solutions to this equation, so the correct answer is (D).

We can rewrite our equation x³ + 2x² − 4x − 8 = 0 using this factorization.

The potential solutions are x = 2 and x = −2. Test each solution in the original rational equation, .

When x = 2:

This numerical equation is true, so 2 is a solution for x.

When x = −2:

The expressions −4/0 and 4/0 are both undefined, so −2 is an extraneous solution.

The only solution to the given rational equation is x = 2.

An alternative means
of factoring the cubic
could have involved
grouping out the
factors in x³ + 2x² − 4x
− 8 = 0, so as to get
x²(x + 2) − 4(x + 2) = 0,
which equals (x² − 4)
(x + 2) = 0. Higher-
degree expressions can
often be manipulated
in this fashion, and
if you can spot these
commonalities, it may
save you some time.

Factoring at the beginning can often help with the solution process. If we had first factored the denominators, we would have gotten the following.

There is a factor of (x + 2) in both denominators. We can multiply both sides of the equation by (x + 2) and cancel out the factor (x + 2) in each denominator, as long as we note that x + 2 ≠ 0, so x ≠ −2. Here is the resulting equation.

where x ≠ −2

When we cross-multiply this equation, we get x² − 2x = 4 − 2x, which simplifies to x² = 4. The two solutions to x² = 4 (which can also be written as x² − 4 = 0) are x = 2 and x = −2. But, we established earlier that x cannot equal −2. That means that the only solution is x = 2. When we test 2 in the original equation, it produces a correct numerical equation, as shown above.

This method is simpler than the first approach we used, involving only a quadratic equation instead of a cubic equation, but it requires making sure the denominator of each rational expression never equals 0.

Because x is another
factor in one of the
denominators, it is
also true that x cannot
equal 0. However, 0
was not a solution
to the polynomial
equation we found
through cross-
multiplying, so it did
not come up as an
extraneous solution.