Solving Rational Inequalities Alternative Format

Lesson Part 1

In This Module

We will employ a variety of methods, both graphical and algebraic, to solve rational inequalities.

Examples

Example 1

Solve the simple rational inequality

The algebraic method of multiplying both sides of an equation by the lowest common denominator to eliminate the denominator, used when solving rational equations, is a more complicated process when dealing with inequalities.

Algebraic Solution

Recall that multiplying both sides of an inequality by a negative value reverses the inequality condition. If we were to multiply both sides of the inequality by \(x-2\), we would need to consider the case when \(x-2 \lt 0\) and the case when \(x-2\geq 0\) in order to have the correct condition on the inequality to move forward:

Case \(1\): \(x-2 \lt 0 \Rightarrow x \lt 2\)

Now, when multiplying both sides by \(x-2\), we must change the condition of the inequality:

Moving forward in solving this inequality,

Since \(x \lt 2\) and \(x \leq 4\), therefore \(x \lt 2\).

Case \(2\): \(x-2 \gt 0 \Rightarrow x \gt 2\)

Now, when multiplying both sides of the inequality by \(x-2\), the inequality condition remains the same:

Since \(x \gt 2\) and \(x \geq 4\), therefore \(x \geq 4\).

We now combine the solution to each of these cases to obtain the overall solution to this inequality.

Therefore, the solution is \(\left\{ x \mid x \lt 2 \right.\) or \(\left. x \geq 4,\ x \in \mathbb{R} \right\}\).

This solution can also be expressed using interval notation, \(x \in (-\infty,2) \cup [4,\infty ),\ x \in \mathbb{R},\) and illustrated using a number line as shown below.

Let's consider a graphic approach to solving this same inequality.

Graphical Solution

To determine the point of intersection of the two functions, set \(f(x)=g(x)\) and solve:

Using the graphs and keeping in mind the asymptote, \(x=2\), and the point of intersection, \((4,3)\), we can determine the intervals in which \(f(x) \leq g(x)\).

Lesson Part 2

Examples

Example 2

Solve \(\dfrac{3}{x+1} \gt -\dfrac{2}{x-2}.\)

Algebraic Solution

First, note the restrictions on \(x\): \(x \neq -1,2\).

Considering the algebraic approach used in the first example, if we were to multiply both sides of the inequality by the lowest common denominator, \((x+1)(x-2)\), we would need to determine intervals when \((x+1)(x-2) \gt 0\) and when \((x+1)(x-2) \lt 0\) to determine the conditions on \(x\) for each case in the solution:

This complicates the procedure, so we will take a different approach:

• Collect all terms to the left side of the inequality and \(0\) to the right side:\[\dfrac{3}{x+1} + \dfrac{2}{x-2} \gt 0\]
• Simplify the expression on the left side by finding a common denominator and adding the terms:\[ \begin{align*} \dfrac{3(x-2)+2(x+1)}{(x+1)(x-2)} & \gt 0 \\ \dfrac{3x-6+2x+2}{(x+1)(x-2)} & \gt 0 \\ \dfrac{5x-4}{(x+1)(x-2)} & \gt 0 \end{align*} \]
• Once simplified, factor the numerator, if possible, always leaving the denominator in factored form. In this case, \(5x-4\) cannot be factored.
• Create an interval table and identify the sign of each factor in the rational expression within each interval, like we did with the polynomial inequalities in an earlier unit.
• The zero value of each factor in the numerator and denominator indicates when a sign change may occur for that factor.

We will need to work with the three zero values, which are \(x=-1, \frac{4}{5}, 2\), and set up a chart as shown. Consider the zeros in numerical order and identify the intervals around these zero values, placing these intervals across the top row. List the factors in the first column. We can then begin identifying the sign of each factor (\(5x-4\), \(x+1\), \(x-2\)) within each interval.

Note: For the full rational function, \(\dfrac{5x-4}{(x+1)(x+2)}\), we consider its three linear factors within each interval. An odd number of negative factors will result in an overall negative value and an even number of negative factors will result in an overall positive value.

Thus, \(\dfrac{5x-4}{(x+1)(x-1)} \gt 0\) when \(-1 \lt x\lt \dfrac{4}{5}\) or \(x \gt 2\).

The solution to the inequality is then \(\left\{x \mid -1 \lt x \lt \dfrac{4}{5} \text{ or } x \gt 2,\ x \in \mathbb{R} \right\}\).

Using interval notation the solution is given by \(x \in (-1, \frac{4}{5}) \cup ( 2, \infty ) , x \in \mathbb{R}\).

Let's solve the same inequality using a graphical approach. This time, we will use technology.

Graphical Solution

To solve the inequality graphically, we will define \(f(x)\) to be equal to the rational expression on the left side of the inequality and \(g(x)\) to the expression on the right side of the inequality. So,

To solve this inequality, we must determine when \(f(x) \gt g(x)\).

Note the point of intersection at \(x=0.8\) and the position of the vertical asymptotes at \(x=-1\) and \(x=2\).

The graph of \(f(x)\), shown in red, is above the graph of \(g(x)\), shown in blue, when \(-1 \lt x \lt 0.8\) or \(x \gt 2\).

Selecting the point of intersection on the graph displays its coordinates in the legend.

The second function is set at \(g(x)=0\), the equation of the \(x\)-axis. The point solution here will be the \(0\) of the function.

Again, keeping in mind the asymptotes, we see from the graph that the solution is

Check Your Understanding A and B

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Lesson Part 3

Examples

Example 3

Solve \(\dfrac{2x}{x+1} \geq \dfrac{x+6}{x+3}\) using an interval sign table.

Algebraic Solution

First, note any restrictions on \(x\): \(x\neq -1, -3\).

Here is a quick review of the steps you should have used.

Take all terms to one side of the inequality:

Simplify that rational expression by finding a common denominator:

In the simplified form, make sure the numerator and denominator are fully factored so you can identify the zero values for those factors to use in the table:

We solve this inequality by using an interval sign table, as before.

Therefore, \(\dfrac{(x-3)(x+2)}{(x+1)(x+3)} \geq 0\) for \(\left\{ x \mid x \lt -3 \text{ or } -2 \leq x \lt -1 \text{ or } x \geq 3,\ x \in \mathbb{R} \right\}\).

Since equality is included in the condition, the zeros of the numerator, \(x=-2\) and \(x=3\), are included in the solution. The zeros of the denominator, \(x=-3\) and \(x=-1\), cannot be included.

Graphical Solution

To solve this inequality graphically without technology would be quite onerous. I would like to take a different graphical approach.

The inequality \(\dfrac{2x}{x+1} \geq \dfrac{x+6}{x+3}\) must first be rearranged and simplified to the form \(\dfrac{(x+3)(x+2)}{(x+1)(x+3)} \geq 0\), as before.

We may now multiply both sides of the inequality by \((x+1)^2(x+3)^2\), which is positive for all real values of \(x,x \neq -1, -3\), since the factors are squared. So there will be no change in the inequality condition when we do this multiplication:

Be careful when doing this, as the simplified version will be a polynomial with no apparent restrictions on \(x\). As mentioned in the module on solving rational equations, we may create extraneous roots or solutions when multiplying both sides of an equation or, in this case, an inequality by an expression of \(x\).

It is important to note the restrictions on \(x\), identified from the original form of the inequality, and keep these in mind when determining the solution. Here the simplified form is a factored quartic polynomial inequality:

We can now solve this polynomial inequality using an interval
table, similar to the one created in the first approach to solving the inequality.

However, sketching polynomials from factored form is fairly straightforward, so we will complete this solution with graphing.

Let \(f(x)=(x+1)(x+3)(x-3)(x+2),\ x \neq -3,-1\).

The function is a quartic polynomial with a positive leading coefficient, i.e., \(1x^4\), if we were to expand out the factored form; therefore, the quartic polynomial will have same end behaviours and so \(y \to \infty\) as \(x \to \pm \infty\).

Thus, \(f(x) \geq 0\) when \(x \lt -3\) or \(-2 \leq x \lt -1 \text{ or } x \geq 3\).

Again, the solution to this inequality is the set

The inadmissible values are not included in the solution.

Check Your Understanding C

This question is not included in the Alternative Format, but can be accessed in the Review section of the side navigation.

Lesson Part 4

Examples

Our fourth example is an application to rational inequalities.

Example 4: Application

An open top box is to be made with the following conditions:

• The length of the base is \(10\text{ cm}\) longer then the width of the base.
• The height of the box must be greater than the width.
• The volume of the box must be \(375\text{ cm}^3\).

Determine specific restrictions, if any, on the width, length, and height of the box.

Solution

Let \(x\) represent the width of the box in centimetres.

The length of the box will be \(x+10\text{ cm}\). Let \(h\) represent the height of the box in centimetres.

Note: \(x \gt 0,\ h \gt 0\).

If the volume is \(375\text{ cm}^3\), then

using the formula

Rearranging the equation \(x(x+10)(h)=375\), we obtain

Now, the height of the box must be greater than the width, so \(h \gt x\).

Substituting the expression for \(h\), we determine the conditions on the width \(x\) such that

We must solve this inequality. Since \(x \gt 0\), then \(x(x+10) \gt 0\). We can then multiply both sides of the inequality by \(x(x+10)\) without changing the inequality condition:

To solve this, we must factor this cubic expression. Use the factor theorem to factor this cubic.

We may use the rational roots theorem to determine test values. These are any integer factors of \(375\).

Observe that if we let \(P(x) = x^3 + 10x^2 - 375\), then \(P(5)=0\) and so \(x-5\) is a factor.

We can find the corresponding quadratic factor by either using long division, synthetic division, or the have and need method:

The first term of the quadratic must be \(x^2\) since \(x(x^2)=x^3\).

The last term must be \(75\) since \(-5(75)=-375\).

Expanding, we have \(x^3-5x^2+75x-375\).

We need \(15x^2-75x=15x(x-5)\).

Thus, the middle term must be \(15x\).

The quadratic factor, \(x^2+15x + 75\), has no real roots since \(b^2-4ac=-75 \lt 0\).

Hence, \((x^2+15x+75) \gt 0\) for all \(x \gt 0\).

Therefore, \(x-5 \lt 0\) since \(\overbrace{(x - 5)}^{\color{BrickRed}-}\overbrace{(x^2 + 15x + 75)}^{\color{BrickRed}+} \lt 0\).

Thus, \(0 \lt x \lt 5\) and \(10 \lt x + 10 \lt 15\).

Therefore, the width of the box must be between \(0\text{ cm}\) and \(5\text{ cm}\), and the length, which is \(10\) centimeters more than the width, must be between \(10\text{ cm}\) and \(15\text{ cm}\).

Now let's find the restrictions on \(h\).

Since \(h=\dfrac{375}{x(x+10)}\) and \(0 \lt x \lt 5\), then \(h \gt \dfrac{375}{5(5+10)}=5\).

The maximum value of the width, \(5\), will generate the minimum value of the height.

This means the height is greater than \(5\), but is ultimately dependent on width.

Summary

• Care must be taken when multiplying both sides of a rational inequality by an expression of \(x\) (usually done to eliminate the denominators). When this expression of \(x\) is negative, the inequality condition must be reversed. More than one case may need to be considered when taking this approach.
• It is often helpful to arrange the terms of the inequality to one side of the condition, with \(0\) on the other side. Once the terms are simplified to a single rational expression in factored form, an interval sign table can be constructed to determine the solution algebraically.
• There are a variety of ways to graphically solve a rational inequality; graphing one or two rational functions or a polynomial function, with or without graphing technology, as demonstrated in this module. Some algebraic steps may be required regardless of the graphical approach employed.