Explainer: One-Variable Quadratic Inequalities

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‰₯<≀greaterthan,greaterthanorequalto,lessthan,lessthanorequalto.

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 square bracket tells us that the 0 is included, and the curved parenthesis 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,∞).

Note that we never put a square bracket next to an infinity symbol as infinity is not a number. 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 square bracket and finish with the infinity symbol with a parenthesis: [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 solve 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‘₯βˆ’3)=0(π‘₯+5)=0.or

So, 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 signs 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, in the intervals π‘₯<βˆ’5 and π‘₯>3. We can express our answer as β„βˆ’[βˆ’5,3].

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 expand the brackets. Note that we should not take the square root of each side here, as this could lead to an invalid answer.

Expanding the brackets gives us (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 inequality has been switched.

Since 24 is a common factor, we can divide all the terms by 24, which gives us π‘₯βˆ’4π‘₯+3β‰₯0.

Setting our 𝑓(π‘₯)=0 and factoring give us π‘₯βˆ’4π‘₯+3=0(π‘₯βˆ’1)(π‘₯βˆ’3)=0.

Therefore, π‘₯βˆ’1=0π‘₯βˆ’3=0π‘₯=1π‘₯=3.oror

To solve the inequality (π‘₯βˆ’1)(π‘₯βˆ’3)β‰₯0, we need to identify the regions where this is true. Whether or not (π‘₯βˆ’1)(π‘₯+3)β‰₯0 depends on the signs of 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. We can write this in interval notation as β„βˆ’(1,3).

As an alternative method, we could have sketched 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. As the roots of the equation are 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. We can express this answer in interval notation as β„βˆ’(1,3).

In the following example, we will look at an alternative method of solving a quadratic inequality by drawing a graph.

Example 4: 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‘₯+21,500 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‘₯+21,500.

Simplifying the terms, we have 𝑃(π‘₯)=βˆ’11π‘₯+1,080π‘₯βˆ’21,500.

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‘₯+1,080π‘₯βˆ’21,500>0.

To solve the inequality, we will sketch a graph of 𝑓(π‘₯)=βˆ’11π‘₯+1,080π‘₯βˆ’21,500. 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‘₯+1,080π‘₯βˆ’21,500=0, we can substitute the values π‘Ž=βˆ’11, 𝑏=1,080, and 𝑐=βˆ’21,500 into the quadratic formula and simplify to get π‘₯=βˆ’1,080±√1,080βˆ’4(βˆ’11)(βˆ’21,500)2(βˆ’11)=βˆ’1,080±√1,166,400βˆ’946,000βˆ’22=βˆ’1,080±√220,400βˆ’22=1,080±√220,40022.

Therefore, π‘₯=1,080+√220,40022π‘₯=1,080βˆ’βˆš220,40022.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‘₯+1,080π‘₯βˆ’21,500 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‘₯+1,080π‘₯βˆ’21,500 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.

Example 5: Solving a Quadratic Inequality Using a Graph

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

Answer

To begin solving this inequality, we will perform the same operation to both of its sides. 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, 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.

Therefore, 2π‘₯βˆ’9=0π‘₯βˆ’3=0.or

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, this is the interval for the inequality. In interval notation, we can write this as [3,4.5].

Note that as the values can also be exactly equal to both 3 and 4.5, we use square brackets beside each value.

Example 6: Solving a Quadratic Inequality Using a Graph

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

Answer

To begin solving this inequality, we will first multiply the parentheses, which gives us (π‘₯βˆ’5)(π‘₯βˆ’7)β‰₯βˆ’5π‘₯+35π‘₯βˆ’12π‘₯+35β‰₯βˆ’5π‘₯+35.

We now need to collect all the terms to the same side of the inequality. Adding 5π‘₯ to both sides, we have π‘₯βˆ’7π‘₯+35β‰₯35.

We can then subtract 35 from both sides, which gives π‘₯βˆ’7π‘₯β‰₯0.

To solve the inequality, we will sketch a graph of 𝑓(π‘₯)=π‘₯βˆ’7π‘₯. To do this, we first need to find the points where the equation crosses the π‘₯-axis. 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‘₯βˆ’7=0,π‘₯=0π‘₯=7.oror

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 π‘₯ is between 0 and 7. We can express this final answer in interval notation as β„βˆ’(0,7).

Key Points

  • We can solve quadratic inequalities by using a similar process to solving quadratic equations, by performing the same arithmetic operations to both sides of the inequality. However, we need to remember that when multiplying or dividing an inequality by a negative number, we must switch the inequality.
  • A quadratic inequality has an interval of solutions, not just one distinct solution.
  • To solve a quadratic inequality algebraically, we follow the steps below:
    1. 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.
    2. Factor the inequality by setting 𝑓(π‘₯)=0 to identify the roots of the expression 𝑓(π‘₯).
    3. 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.
  • Take special care if you have rearranged the inequality to change the sign of the π‘₯ value; use the rearranged form of the inequality to identify the intervals, rather than the original inequality.
  • To solve a quadratic inequality graphically, we follow the steps below.
    1. 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.
    2. Factor the inequality by setting 𝑓(π‘₯)=0, to identify the roots of the expression 𝑓(π‘₯).
    3. 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.
    4. Identify the intervals that satisfy the inequality.
  • Our drawn graph does not need to be precise or even look particularly neat; it is simply there to give us a visual picture so that we can reach a conclusion about the intervals where the inequality will be true.

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