Lesson Explainer: One-Variable Quadratic Inequalities Mathematics

In this explainer, we will learn how to solve one-variable quadratic inequalities algebraically and graphically.

Recall that, in an equation, we have two expressions that are equal to each other, and we write the equals sign, =, between them. When we have two expressions that are not equal to one another, we can relate the expressions by the use of an inequality sign.

We can have inequalities such as π‘₯β‰₯4,6≀π‘₯,2π‘₯βˆ’7>5.

In each of these inequalities, π‘₯ has a range of possible solutions. When we have an inequality such as π‘₯β‰₯4, we can say this in words as β€œπ‘₯ is greater than or equal to four.” This means that a value of π‘₯ that is four or more will satisfy this inequality. The four inequality symbols we use are >,β‰₯,<,≀.greaterthangreaterthanorequaltolessthanlessthanorequalto

We can solve inequalities in a similar process to solving equations, by ensuring that we perform the same mathematical operation to both sides of the inequality. However, as inequalities have a direction, we must be careful to consider which side of the inequality an expression is on. When we multiply or divide by a negative number, we must switch the inequality. For example, if we have βˆ’π‘₯β‰€βˆ’2, then when dividing by βˆ’1 we must switch the inequality to give π‘₯β‰₯2.

Let us now look at how to solve an inequality and represent the answer as an interval. Before doing so, we need to recap some notation. If we consider the interval of numbers from 0 to 10, which includes 0 but not 10, we could represent this using inequalities as 0≀π‘₯<10.

The strict inequality on the right tells us that the 10 is not included in the inequality, and the nonstrict inequality on the left tells us that the 0 is included. Another way of writing this interval would be [0,10[.

Here, the closed square bracket tells us that the 0 is included, and the open square bracket tells us that the 10 is not included. It is also worth recapping here that the symbol for infinity is ∞. This is often used to represent intervals that are greater than or less than a single number. For example, π‘₯>3 in interval notation would be ]3,∞[.

We will now look at an example of solving a linear inequality.

Example 1: Finding the Solution Set of a Linear Inequality

Find the solution set of the inequality βˆ’2π‘₯+3≀5. Write your answer as an interval.

Answer

First, we subtract 3 from both sides of the inequality βˆ’2π‘₯+3≀5, which gives us βˆ’2π‘₯≀5βˆ’3βˆ’2π‘₯≀2.

We can now divide each side by βˆ’2, remembering that when we divide an inequality by a negative number, we need to switch the inequality sign. This gives us π‘₯β‰₯βˆ’1.

So π‘₯ is all the numbers greater than or equal to βˆ’1.

To express the solution as an interval, we must represent all the numbers from βˆ’1 up to infinity, including βˆ’1. So, we start our interval with a closed square bracket and finish with the infinity symbol with open square bracket: [βˆ’1,∞[.

In the same way that we have distinct equations such as linear and quadratic equations, we can have quadratic inequalities in the following forms.

Definition: Quadratic Inequality

A quadratic inequality can be in one of the following forms: π‘Žπ‘₯+𝑏π‘₯+𝑐>0,π‘Žπ‘₯+𝑏π‘₯+𝑐β‰₯0,π‘Žπ‘₯+𝑏π‘₯+𝑐<0,π‘Žπ‘₯+𝑏π‘₯+𝑐≀0, where π‘Ž, 𝑏, and 𝑐 are constants and π‘Žβ‰ 0.

When we solve a quadratic inequality, we need to find the range of solutions, or intervals, for which an inequality is true. We can solve quadratic inequalities using the process steps below.

How To: Solving a Quadratic Inequality Algebraically

  1. Rearrange the inequality so that we have all the terms of the expression, defined as 𝑓(π‘₯), on one side, with an inequality relating this to zero. For example, 𝑓(π‘₯)≀0 or 𝑓(π‘₯)>0.
  2. Solve 𝑓(π‘₯)=0 by factoring, or otherwise, to find the solutions to the equation.
  3. Select test points for each interval so that there are values less than, between, and greater than the solutions of the equation. We can also use a sign chart to identify the intervals that will be positive or negative.
  4. Identify the intervals that satisfy the inequality.

Let us now look at an example of solving a quadratic inequality using test points to identify the interval solution.

In the next example, we will look at how we can use a sign chart to identify positive and negative values for the intervals of an inequality.

Example 2: Solving a Quadratic Inequality Using a Sign Chart

Describe all solutions to the inequality 15βˆ’π‘₯βˆ’2π‘₯<0.

Answer

To begin solving the inequality 15βˆ’π‘₯βˆ’2π‘₯<0, we will first transform and rearrange this to get a positive coefficient of π‘₯. We can multiply all terms of the coefficient by βˆ’1, recalling that when we multiply an inequality by a negative number, we must switch the inequality. This gives us 15βˆ’π‘₯βˆ’2π‘₯<0π‘₯+2π‘₯βˆ’15>0.

We now need to solve 𝑓(π‘₯)=0, where 𝑓(π‘₯)=π‘₯+2π‘₯βˆ’15. We can factor our equation to give π‘₯+2π‘₯βˆ’15=0(π‘₯βˆ’3)(π‘₯+5)=0.

Therefore, π‘₯=3π‘₯=βˆ’5.or

To solve the inequality (π‘₯βˆ’3)(π‘₯+5)>0, we need to identify the regions where this is true. Whether or not (π‘₯βˆ’3)(π‘₯+5)>0 depends on the signs of the factors (π‘₯βˆ’3) and (π‘₯+5).

We can create a grid to identify whether each factor will be positive or negative in the intervals less than, greater than, and between our solutions of π‘₯=βˆ’5 and π‘₯=3. In the grid, we can place the intervals between our solutions horizontally and our factors of 𝑓(π‘₯) vertically, with the product of the factors below. We can then calculate whether the product of the factors will be positive or negative.

π‘₯<βˆ’5βˆ’5<π‘₯<3π‘₯>3
(π‘₯βˆ’3)βˆ’βˆ’+
(π‘₯+5)βˆ’++
(π‘₯βˆ’3)(π‘₯+5)+βˆ’+

In the grid above, we can see in the first results column that when π‘₯<βˆ’5, our values (π‘₯βˆ’3) and (π‘₯+5) will both be negative, so the product of those two negative values, (π‘₯βˆ’3)(π‘₯+5), will be positive.

Checking the sign of (π‘₯βˆ’3)(π‘₯+5) in the grid, we can see that it will be positive, that is, (π‘₯βˆ’3)(π‘₯+5)>0, when π‘₯<βˆ’5 or π‘₯>3. In other words, the inequality 15βˆ’π‘₯βˆ’2π‘₯<0 is satisfied when π‘₯ does not satisfy βˆ’5≀π‘₯≀3. In the interval notation, we can express our answer as β„βˆ’[βˆ’5,3].

In the previous example, we solved a quadratic equation where the right side of the inequality is equal to zero. When both sides of the inequality contain nonzero expressions, we need to first simplify the inequality to the point where one side of it is zero. In the next example, we will consider an example where both sides of the inequality contain quadratic expressions. After simplifying this inequality, we will solve it both algebraically and graphically.

Example 3: Solving a Quadratic Inequality

Determine the solution set of the inequality (π‘₯+3)≀(5π‘₯βˆ’9).

Answer

To begin solving this inequality, we will first simplify it so that one side is zero. Although it is tempting, we cannot simply take the square root of each side of the equality here. Since the square root can take both positive and negative values, taking a square root of an inequality can lead to an invalid answer. Instead, we can multiply through the parenthesis in both sides of the inequality to obtain (π‘₯+3)≀(5π‘₯βˆ’9)π‘₯+6π‘₯+9≀25π‘₯βˆ’90π‘₯+81.

We now need to collect all the terms on the same side of the inequality. In order to keep a positive coefficient of π‘₯, we can subtract all the terms on the left-hand side from each side of the inequality, which gives us 0≀25π‘₯βˆ’90π‘₯+81βˆ’ο€Ήπ‘₯+6π‘₯+9.

Simplifying, we have 0≀24π‘₯βˆ’96π‘₯+72.

We can also write this as 24π‘₯βˆ’96π‘₯+72β‰₯0, noting that the two sides of the inequality have been switched.

Since 24 is a common factor, we can divide all the terms by 24, which gives us

π‘₯βˆ’4π‘₯+3β‰₯0.(1)

We have simplified the given inequality to the point where one side of it is zero. Now, we will finish solving this inequality using two different methods: the algebraic method and the graphical method.

Method 1

We first solve this inequality algebraically. Setting our 𝑓(π‘₯)=0 and factoring give us π‘₯βˆ’4π‘₯+3=0(π‘₯βˆ’1)(π‘₯βˆ’3)=0.

Therefore, π‘₯=1π‘₯=3.or

In factorized form, inequality (1) is written as (π‘₯βˆ’1)(π‘₯βˆ’3)β‰₯0, and we need to identify the regions where this inequality is true. Whether or not (π‘₯βˆ’1)(π‘₯+3)β‰₯0 depends on the factors (π‘₯βˆ’1) and (π‘₯βˆ’3).

We can create a grid to identify whether each factor will be positive or negative in the intervals less than, greater than, and between our solutions π‘₯=1 and π‘₯=3. Since we have a nonstrict inequality, we can also record the values when π‘₯=1 and π‘₯=3. We can then calculate whether the product of the factors will be positive or negative.

π‘₯<1π‘₯=11<π‘₯<3π‘₯=3π‘₯>3
(π‘₯βˆ’1)βˆ’0+++
(π‘₯βˆ’3)βˆ’βˆ’βˆ’0+
(π‘₯βˆ’1)(π‘₯βˆ’3)+0βˆ’0+

From the grid, we can see that the intervals where (π‘₯βˆ’1)(π‘₯βˆ’3)β‰₯0 are when π‘₯≀1 and when π‘₯β‰₯3. In other words, the inequality (π‘₯+3)≀(5π‘₯βˆ’9) is satisfied when π‘₯ does not satisfy 1<π‘₯<3. In the interval notation, we can express our answer as β„βˆ’]1,3[.

Method 2

Let’s consider how to solve inequality (1) graphically. We begin by sketching a graph of 𝑓(π‘₯)=π‘₯βˆ’4π‘₯+3. Given that the coefficient of π‘₯ is positive, we know that the parabola curve will open upward. In method 1, we have identified the roots of the equation to be π‘₯=1 and π‘₯=3, this means that the curve will pass through the coordinates (1,0) and (3,0).

To solve the inequality π‘₯βˆ’4π‘₯+3β‰₯0, we consider the points on the graph of 𝑓(π‘₯)=π‘₯βˆ’4π‘₯+3, where 𝑓(π‘₯)β‰₯0. This will be above the π‘₯-axis, at the values where π‘₯≀1 and where π‘₯β‰₯3. As shown in method 1, our answer in interval notation can be written as β„βˆ’]1,3[.

In the previous example, we solved a quadratic inequality both algebraically and graphically. Both methods required simplifying the given inequality to the point where one side of it is zero. From there, it is simple to solve a quadratic inequality graphically as long as we can sketch the graph of the quadratic function. We will use the graphical method of solving quadratic inequalities in the remaining examples.

Let us consider another example of solving a quadratic inequality graphically.

Example 4: Solving a Quadratic Inequality Using a Graph

Solve the inequality 2π‘₯≀15π‘₯βˆ’27.

Answer

We begin by simplifying the given inequality to the point where one side of it is zero. We can subtract 15π‘₯ from both sides, which gives us 2π‘₯≀15π‘₯βˆ’272π‘₯βˆ’15π‘₯β‰€βˆ’27.

We can then add 27 to both sides of the inequality, giving us 2π‘₯βˆ’15π‘₯+27≀0.

To solve the inequality graphically, we will sketch a graph of 𝑓(π‘₯)=2π‘₯βˆ’15π‘₯+27. To do this, we first need to find the points where the equation crosses the π‘₯-axis, often called the roots of the equation.

Setting our 𝑓(π‘₯)=0, we can factor this, giving us 2π‘₯βˆ’15π‘₯+27=0(2π‘₯βˆ’9)(π‘₯βˆ’3)=0.

So, π‘₯=4.5π‘₯=3.or

We now need to establish the shape of the curve 𝑓(π‘₯)=2π‘₯βˆ’15π‘₯+27. As the coefficient of π‘₯, 2, is positive, this means that the parabola curve will open upward.

So, as the roots of the equation are π‘₯=4.5 and π‘₯=3, we can plot the coordinates (4.5,0) and (3,0) and sketch a parabola curve as shown below.

Next, we need to identify the areas for which the inequality 2π‘₯βˆ’15π‘₯+27≀0 holds true. We can see from the sketch that 𝑓(π‘₯)=2π‘₯βˆ’15π‘₯+27 is at values less than zero between the values π‘₯=3 and π‘₯=4.5. Therefore, π‘₯ must satisfy 3≀π‘₯≀4.5. In interval notation, we can write this as [3,4.5].

We consider one more example of solving a quadratic inequality graphically.

Example 5: Solving a Quadratic Inequality Using a Graph

Solve the inequality (π‘₯βˆ’5)(π‘₯βˆ’7)β‰₯βˆ’5π‘₯+35.

Answer

We begin by simplifying the given inequality to the point where one side of it is zero. Multiplying through the parenthesis, we obtain (π‘₯βˆ’5)(π‘₯βˆ’7)β‰₯βˆ’5π‘₯+35π‘₯βˆ’12π‘₯+35β‰₯βˆ’5π‘₯+35π‘₯βˆ’7π‘₯+35β‰₯35π‘₯βˆ’7π‘₯β‰₯0.

To solve the inequality graphically, we will sketch a graph of 𝑓(π‘₯)=π‘₯βˆ’7π‘₯. To do this, we first need to find the roots of the quadratic function 𝑓(π‘₯). These roots can be found by setting 𝑓(π‘₯)=0 and solving, giving us π‘₯βˆ’7π‘₯=0.

Factoring, we have π‘₯(π‘₯βˆ’7)=0.

Therefore, π‘₯=0π‘₯=7.or

As the coefficient of π‘₯ in the equation 𝑓(π‘₯)=π‘₯βˆ’7π‘₯ is 1, this value is greater than zero; so the parabola curve will open upward. As the roots of the equation are π‘₯=0 and π‘₯=7, this means that the curve will pass through the coordinates (0,0) and (7,0). We can sketch the graph shown below.

To solve the inequality π‘₯βˆ’7π‘₯β‰₯0, we consider the points on the graph of 𝑓(π‘₯)=π‘₯βˆ’7π‘₯, where 𝑓(π‘₯)β‰₯0. This will be above the π‘₯-axis at the values where π‘₯ is less than 0 and where π‘₯ is greater than 7. Because we do not have a strict inequality, π‘₯ can also be exactly equal to 0 or 7. Another way to express this would be by saying that π‘₯ is all the values excluding the points where 0<π‘₯<7. We can express this final answer in interval notation as β„βˆ’(0,7).

In our final example, we will solve a quadratic inequality in a real-world business application.

Example 6: Solving a Quadratic Inequality in a Real-World Context

A cell phone company has the following cost and revenue functions: 𝐢(π‘₯)=8π‘₯βˆ’600π‘₯+21500 and 𝑅(π‘₯)=βˆ’3π‘₯+480π‘₯, where π‘₯ is the number of cell phones.

State the range for the number of cell phones they can produce while making a profit. Round your answers to the nearest integer that guarantees a profit.

Answer

In beginning of this question, we need to apply some real-world mathematics. We can calculate profit using the following calculation: profitrevenuecost=βˆ’.

We can define the profit as 𝑃(π‘₯), and substituting the given functions of cost, 𝐢(π‘₯), and revenue, 𝑅(π‘₯), we have 𝑃(π‘₯)=𝑅(π‘₯)βˆ’πΆ(π‘₯)𝑃(π‘₯)=βˆ’3π‘₯+480π‘₯βˆ’ο€Ή8π‘₯βˆ’600π‘₯+21500.

Simplifying the terms, we have 𝑃(π‘₯)=βˆ’11π‘₯+1080π‘₯βˆ’21500.

We are asked to find the range for which a profit can be made. This would be a profit greater than 0, which we can write as the inequality βˆ’11π‘₯+1080π‘₯βˆ’21500>0.

To solve the inequality graphically, we will sketch a graph of 𝑓(π‘₯)=βˆ’11π‘₯+1080π‘₯βˆ’21500. To do this, we first need to solve the equation 𝑓(π‘₯)=0, to find the points where the equation crosses the π‘₯-axis. To solve, we will use the quadratic formula.

Recall that a quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are constants, can be solved using the quadratic formula π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Therefore, to solve the equation βˆ’11π‘₯+1080π‘₯βˆ’21500=0, we can substitute the values π‘Ž=βˆ’11, 𝑏=1080, and 𝑐=βˆ’21500 into the quadratic formula and simplify to get π‘₯=βˆ’1080±√1080βˆ’4(βˆ’11)(βˆ’21500)2(βˆ’11)=βˆ’1080±√1166400βˆ’946000βˆ’22=βˆ’1080±√220400βˆ’22=1080±√22040022.

Therefore, π‘₯=1080+√22040022π‘₯=1080βˆ’βˆš22040022.or

Using a calculator, we can evaluate π‘₯ to one decimal place, which gives us π‘₯=70.4π‘₯=27.8.or

Next, we can sketch a graph, plotting the coordinates of the roots at (27.8,0) and (70.4,0). Note that as this is just a sketch to help illustrate the inequality, it does not need to be precise. Since the equation 𝑓(π‘₯)=βˆ’11π‘₯+1080π‘₯βˆ’21500 has a coefficient of π‘₯ (βˆ’11) that is less than zero, the equation will have a curve that opens downward. Hence, the graph will look as below.

Next we will need to establish the points where 𝑓(π‘₯)>0. The curve 𝑓(π‘₯)=βˆ’11π‘₯+1080π‘₯βˆ’21500 has values greater than zero between the π‘₯-values of 27.8 and 70.4. This means that the range of cell phones that can be produced while making a profit is between 27.8 and 70.4 cells. However, as we need to give our answers to the nearest integer, our final answer is 28–70 cell phones.

Let’s finish by recapping a few important concepts from the explainer.

Key Points

  • The solution of a quadratic inequality is an interval or a union of intervals. When it is a union of two intervals, we can use the set difference notation to write it as the complement of one interval.
  • To solve a quadratic inequality algebraically, we follow the steps below:
    • Rearrange the inequality so that we have all the terms of the expression on one side, with an inequality relating this to zero, for example, 𝑓(π‘₯)>0.
    • Factor the inequality by setting 𝑓(π‘₯)=0 to identify the roots of the expression 𝑓(π‘₯).
    • Identify the intervals that satisfy the inequality by using test points in each interval or a sign chart. We can also sketch a graph of the function.
  • To solve a quadratic inequality graphically, we follow the steps below.
    • Rearrange the inequality so that we have all the terms of the expression on one side, with an inequality relating this to zero; for example 𝑓(π‘₯)>0.
    • Factor the inequality by setting 𝑓(π‘₯)=0, to identify the roots of the expression 𝑓(π‘₯).
    • Sketch the graph of the equation 𝑓(π‘₯)=0, using the roots of the equation and finding the direction of the parabola curve. Take special care if you have rearranged the original inequality to change the sign of the π‘₯ value: use the π‘₯ coefficient in the rearranged form of the inequality to identify the shape of the curve, rather than the original inequality.
    • Identify the intervals that satisfy the inequality.

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