Lesson Explainer: Solving Quadratic Equations Using Graphs Mathematics

In this explainer, we will learn how to graph a quadratic function of the form 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, in order to solve the equation 𝑓(π‘₯)=0.

When working with any particular quadratic function, it can be helpful to remind ourselves of some basic properties that these functions hold, as well as the key features that can be used to classify or describe them. Consider the general quadratic function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+𝑐, where π‘Ž, 𝑏, and 𝑐 are real numbers and π‘Žβ‰ 0. When plotting this function, there will be two basic shapes that we will see, depending on the sign of the parameter π‘Ž. If π‘Ž is positive, then the plot of the function will have a β€œu” shape, and if π‘Ž is negative, then the plot of the function will have an β€œn” shape. We demonstrate this on the graphs below, where the left-hand graph is the plot of a quadratic function where π‘Ž is positive, and the right-hand graph is the plot of a quadratic function where π‘Ž is negative. We have deliberately omitted the π‘₯𝑦-axis, as well as the gridlines.

The values of the other two parameters, 𝑏 and 𝑐, will of course affect the graph when plotting a quadratic function, and one way that this is understood is by classifying how all parameters influence the roots of the function, which we will focus on later. However, the main focus of this explainer will be to understand how roots of a quadratic function can be found by plotting the graph. This actually splits the problem into three distinct categories, each of which we will give a short example of. Before doing so, we will give a definition of the so-called solution set of a quadratic, as follows.

Definition: The Solution Set of a Quadratic Function

Consider the quadratic function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+𝑐, where π‘Ž, 𝑏, and 𝑐 are real numbers and π‘Žβ‰ 0. Then the solution set 𝑆 of the function 𝑓(π‘₯) consists of all possible roots of 𝑓(π‘₯). In other words, 𝑆 is the set of all values of π‘₯ such that 𝑓(π‘₯)=0, which is where the graph crosses the π‘₯-axis. There are three distinct possibilities for the solution set:

  • There are two real roots of 𝑓(π‘₯), when π‘₯=π‘ οŠ§ and π‘₯=π‘ οŠ¨, where π‘ β‰ π‘ οŠ§οŠ¨. In this case, the solution set 𝑆 has two elements and is written 𝑆={𝑠,𝑠}.
  • There is one real root of 𝑓(π‘₯), when π‘₯=π‘ οŠ§. In this case, the solution set has one element and is written 𝑆={𝑠}.
  • There are no real roots of 𝑓(π‘₯), meaning that the solution set is empty. In this case, we say that 𝑆 is equal to the β€œempty set” which is denoted 𝑆=βˆ….

Consider the quadratic function 𝑓(π‘₯)=π‘₯βˆ’π‘₯βˆ’2. One excellent way of understanding the behavior of this function (or any function) is to plot a graph or a high-quality sketch. We can begin doing this by plotting a table of values and then converting this information to coordinates in the π‘₯𝑦-plane. For the quadratic function 𝑓(π‘₯)=π‘₯βˆ’π‘₯βˆ’2, we currently have no knowledge of how this behaves and therefore we do not have any information as to where interesting behavior might occur (such as the location of the roots). With no better guess at hand, we will create a table of values for the few points near to the π‘₯-axis. In particular, we will create a table of values for π‘₯=βˆ’3,βˆ’2,βˆ’1,0,1,2,3, which in this case gives the following.

π‘₯βˆ’3βˆ’2βˆ’10123
𝑓(π‘₯)1040βˆ’2βˆ’204

Now that we have this information about the function, we can plot seven coordinates to better gain an understanding of its behavior. We would begin with the coordinate (βˆ’3,10), then with (βˆ’2,4) and so on until we had obtained the plot below.

This information alone offers a much better understanding of how the function behaves. Either from the table of values above, or from the plotted coordinates, we can see that there are two points where the function meets with the π‘₯-axis. It is perhaps clearer to first plot the full quadratic function in this region, which we have done below.

In the graph above, we have marked the points where the function crosses the π‘₯-axis in red crosses. This appears to occur in two locations: when π‘₯=βˆ’1 and π‘₯=2. Equally, we could say that the coordinates of the roots are (βˆ’1,0) and (2,0). Finally, we could use the language of the solution set as described in the definition above, and we would say the solution set is 𝑆={βˆ’1,2}. If we wished to, we could check that these were the correct roots by factoring 𝑓(π‘₯), which in this case would return 𝑓(π‘₯)=(π‘₯+1)(π‘₯βˆ’2). The roots of the function occur when 𝑓(π‘₯)=0, which is precisely when π‘₯+1=0 or when π‘₯βˆ’2=0; that is, when π‘₯=βˆ’1 or π‘₯=2.

Now, we will give an example of a quadratic function where there is only one root, and we will see how this invokes a curios property that will appear several times again in this explainer. Suppose that we now had the quadratic function 𝑓(π‘₯)=4π‘₯βˆ’32π‘₯+64. A superb first step toward understanding this function is with the following table of values.

π‘₯2345678
𝑓(π‘₯)16404163664

Unlike the previous example, every value of 𝑓(π‘₯) is zero or positive, which might hint that there is a fundamentally different behavior occurring. In the graph below, we have plotted the table of values as well as the quadratic function within this region. We have had to scale the axis in order to properly demonstrate the behavior around the π‘₯-axis, which is where any roots will occur.

There is clearly one root when π‘₯=4, which is marked with a red cross. Alternatively, we could say that the single root has the coordinate (4,0), or that the solution set is 𝑆={4}. Notice how the single root is also the minimum point of the quadratic function. This is an instance of a more general result: when the solution set has one element, the vertex maximumminimumpoint is also the only root of the function. Had we chosen to factor the quadratic then, we would have found that 𝑓(π‘₯)=(2π‘₯βˆ’8) and then 𝑓(π‘₯)=0 occurs only when π‘₯=4.

For our final example, we will consider the quadratic function 𝑓(π‘₯)=βˆ’π‘₯βˆ’3π‘₯βˆ’5, which produces the table of values below. We should note that all given values of 𝑓(π‘₯) are negative and, crucially, there are none that are zero, which is a clue that the behavior is different from the previous example.

π‘₯βˆ’3βˆ’2βˆ’10123
𝑓(π‘₯)βˆ’5βˆ’3βˆ’3βˆ’5βˆ’9βˆ’15βˆ’23

We have plotted this function below and can see immediately that there are no points where the function crosses the π‘₯-axis, meaning that there are no roots of the function. Given this, we would write the solution set as 𝑆=βˆ…. Had we attempted to factor this quadratic, either by completing the square or through other means, then we would have found that this was not possible as, at some point, we would have been attempting to take the square root of a negative number. We will comment more on this later in the explainer.

Now we have seen an example of solution set with two elements, a solution set with one element, and a solution set with zero elements. There are various relationships between parameters that can be established which determine critical behaviors of quadratic functions. We will allude to several such examples throughout the rest of this explainer, but briefly we will reconsider the function plotted immediately above. Notice that the vertex (the maximum point) is below the π‘₯-axis. Given that the plot of this quadratic is an β€œn” shape, this means that there cannot be any points where the function crosses the π‘₯-axis because, by definition, a function cannot ever exceed its maximum value. It is possible to argue such results with total algebraic rigor, which we will leave as an exercise for the interest of the reader.

We could also be asked a question where we are required to plot the graph of a quadratic and then use this to estimate its roots (and therefore the solution set). As ever, it is generally good practice to plot or sketch the graph of any function, no matter what the intended purpose of working with this function actually is. Often, this will give a strong indication as to the behavior of the function and the nature of any specific features (such as roots and turning points).

Let us consider the function 𝑓(π‘₯)=3π‘₯βˆ’3π‘₯βˆ’18 in the interval π‘₯=βˆ’4 to π‘₯=4 for integer values of π‘₯. By plotting the graph of the function, we can then use this to estimate the solutions to equation 𝑓(π‘₯)=0. We create the following table of values for the function.

π‘₯βˆ’4βˆ’3βˆ’2βˆ’101234
𝑓(π‘₯)42180βˆ’12βˆ’18βˆ’18βˆ’12018

We can use this to plot the graph as follows.

In order to best demonstrate the features around the line 𝑓(π‘₯)=0, we can scale the axes accordingly. As we can see, this graph seems to strongly indicate that the the roots of 𝑓(π‘₯) are found when π‘₯=βˆ’2 and π‘₯=3, which would imply that the solution set can be written as 𝑆={βˆ’2,3}. In fact, in this case, it is obvious from the table of values, since we clearly see that 𝑓(βˆ’2)=𝑓(3)=0.

In our first formal example, we will enjoy the luxury of having the graph preplotted for our convenience. This will not always be the case, and in real life we would often have to plot such a curve manually or, even better, use an online graph platform to gain a more precise (and speedier) understanding.

Example 1: Finding the Solution Set of a Quadratic Equation Graphically

The diagram shows the graph of 𝑦=𝑓(π‘₯). What is the solution set of the equation 𝑓(π‘₯)=0?

Answer

From this graph, we can see that there is only one root of this quadratic, which occurs when π‘₯=βˆ’2. The coordinate of this root is (βˆ’2,0), which is also the location of the minimum point of the function. The solution set is 𝑆={βˆ’2}.

Although it was not stated in the previous example, the quadratic function was actually 𝑓(π‘₯)=(π‘₯+2)=π‘₯+4π‘₯+4. Since we know that the root of this function is when π‘₯=βˆ’2, we can create a table of values in this region.

π‘₯βˆ’5βˆ’4βˆ’3βˆ’2βˆ’101
𝑓(π‘₯)9410149

On the graph below, we have plotted these particular points and have marked the single root in red. The π‘₯-axis has also been highlighted in red and we can see that this intersects the function 𝑓(π‘₯) precisely at the point (βˆ’2,0).

Example 2: Finding the Solution Set of a Quadratic Equation Graphically

The diagram shows the graph of 𝑦=𝑓(π‘₯). What is the solution set of the equation 𝑓(π‘₯)=0?

Answer

The graph appears to cross the π‘₯-axis when π‘₯=βˆ’3 and π‘₯=1, or alternatively at the points (βˆ’3,0) and (1,0). This means that the solution set has two elements, being written in full as 𝑆={βˆ’3,1}.

Example 3: Finding the Solution Sets of Quadratic Equations from Their Graphs

Consider the following graph of 𝑦=𝑓(π‘₯). What is the solution set of the equation 𝑓(π‘₯)=0?

Answer

This quadratic function clearly has two roots that appear to be when π‘₯=βˆ’2 and π‘₯=2, or alternatively at the coordinates (βˆ’2,0) and (2,0). The solution set therefore has two elements and is written as 𝑆={βˆ’2,2}.

Example 4: Identifying the Solution Set of a Quadratic given Its Graph

The graph shows the function 𝑓(π‘₯)=π‘₯βˆ’2π‘₯+3. What is the solution set of 𝑓(π‘₯)=0?

Answer

There are no points where the graph crosses the π‘₯-axis, meaning that there are no roots of 𝑓(π‘₯). Therefore, the solution set is 𝑆=βˆ….

Although the question does not require us to demonstrate that 𝑓(π‘₯) has no solutions, we will give a quick explanation. Suppose that we tried to find the roots of 𝑓(π‘₯) by completing the square. Then, we can write 𝑓(π‘₯) as follows: 𝑓(π‘₯)=π‘₯βˆ’2π‘₯+3=(π‘₯βˆ’1)βˆ’1+3=(π‘₯βˆ’1)+2.

If we were not trying to find the points where 𝑓(π‘₯)=0, then we would be attempting to solve the equation (π‘₯βˆ’1)+2=0.

This is equivalent to requiring that (π‘₯βˆ’1)=βˆ’2, and it is at this point that it is clear why there are no real values of π‘₯ that allow this. If we were to try solving the above equation for π‘₯, then we would first have to eliminate the squared term on the left-hand side by taking the square root of both sides. However, this is not possible given that the right-hand side is negative and we cannot take the square root of a negative number. For this reason, there are no real values of π‘₯ that solve the above equation and hence there are no solutions to the equation 𝑓(π‘₯)=0. Stated slightly differently, there are no roots of 𝑓(π‘₯) and hence the associated solution set 𝑆 is empty.

In the examples above, we were given the graph of different quadratic functions 𝑓(π‘₯), which allowed us to visually identify the points where these graphs crossed the π‘₯-axis. This information allowed us to write these roots of the function in terms of the associated solution set 𝑆, which could have either two distinct elements, one single element, or no elements at all. In the case of there only being a single element, the location of the root would be common with the only vertex of the function.

Even though we were provided with the graphs of the quadratic functions, this information is not necessary if we are only interested in writing the solution set. Helpful information in determining the solution set can be as follows: the precise coordinates of the roots; the sign of the leading coefficients (which determines whether the graph is a β€œu” shape or an β€œn” shape); and the location of the vertex. It is important to note that knowing this information is not the same as knowing each of the coefficients π‘Ž, 𝑏, or 𝑐 for a quadratic 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨. To clarify, just because we know the solution set of a quadratic function, it does not mean that we know what the function is. This is because two or more quadratic functions can have the same solution set, meaning that complete knowledge of the solution set is not enough to know the quadratic function itself (although, it does tell us most of the information that we would require).

As an example, consider the two quadratic functions 𝑓(π‘₯)=π‘₯+3π‘₯βˆ’4,𝑓(π‘₯)=βˆ’2π‘₯βˆ’6π‘₯+8.

These two functions are plotted below, with 𝑓(π‘₯) being plotted in red and 𝑓(π‘₯) plotted in blue. We can see that they both have the solution set 𝑆={βˆ’4,1} due to the roots which are marked at (βˆ’4,0) and (1,0) respectively. Not only are these functions clearly different, but they have a fundamentally different shape, which is due to the different sign on the leading coefficient of each function. Actually, the two functions are related by the formula 𝑓(π‘₯)=βˆ’2𝑓(π‘₯), which explains why these two functions share a solution set, since the roots that populate this set must solve the equation 𝑓(π‘₯)=βˆ’2𝑓(π‘₯)=0. More generally, any two quadratics 𝑓(π‘₯)=π‘˜π‘“(π‘₯) will have an identical solution set for any constant π‘˜β‰ 0. This reiterates the earlier statement that knowledge of the solution set is not enough to fully describe the quadratic function itself.

Although it is above the level of this explainer, it can be helpful to be mindful of the following classification method for the solution set. For the general quadratic function, the roots can be written in terms of the (mathematically) famous quadratic root formula: π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

The values of π‘Ž, 𝑏, and 𝑐 will not only affect the values of the roots, but also decide the number of roots that the quadratic function has (and hence the type of solution set). In the quadratic root formula above, we have the expression βˆšπ‘βˆ’4π‘Žπ‘οŠ¨. If it is the case that 𝑏<4π‘Žπ‘οŠ¨, then the term within the square root function is negative, meaning that we are trying to take the square root of a negative number. This is not possible when working only with the real numbers, meaning that a general quadratic may not have any roots. If it is the case that 𝑏=4π‘Žπ‘οŠ¨, then we will be taking the square root of zero and hence the Β± sign will be irrelevant, meaning that there is a single root when π‘₯=βˆ’π‘2π‘Ž. In the case that 𝑏>4π‘Žπ‘οŠ¨, we will be taking the square root of a positive number, which will return two possible solutions from the Β± operator. Although outside the scope of this explainer, it can be helpful to use the quadratic root formula to classify these.

Definition: The Solution Set of a Quadratic Function, Giving Reference to the Quadratic Equation

Consider the quadratic function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+𝑐, where π‘Ž, 𝑏, and 𝑐 are real numbers and π‘Žβ‰ 0. Then, the solution set 𝑆 of the function 𝑓(π‘₯) consists of all possible roots of 𝑓(π‘₯). In other words, 𝑆 is the set of all values of π‘₯ such that 𝑓(π‘₯)=0. The solution set can therefore be classified as follows:

  • 𝑏>4π‘Žπ‘οŠ¨; then 𝑆=ο°βˆ’π‘+βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž,βˆ’π‘βˆ’βˆšπ‘βˆ’4π‘Žπ‘2π‘ŽοΌοŠ¨οŠ¨;
  • 𝑏=4π‘Žπ‘οŠ¨; then 𝑆=ο­βˆ’π‘2π‘ŽοΉ;
  • 𝑏<4π‘Žπ‘οŠ¨; then 𝑆=βˆ….

The symbol βˆ… represents the set with no elements, known usually as the β€œempty set.”

A less precise but simpler way of understanding this is as follows. Assume that there is a quadratic function 𝑓(π‘₯) with the solution set 𝑆={𝑠,𝑠}, where π‘ β‰ π‘ οŠ§οŠ¨. In this case, it is clear that the function 𝑓(π‘₯)=(π‘₯βˆ’π‘ )(π‘₯βˆ’π‘ ) has the given solution set 𝑆, since 𝑓(𝑠)=𝑓(𝑠)=0. It is also the case that any function 𝑔(π‘₯)=π‘˜π‘“(π‘₯) will have the solution set 𝑆 for any π‘˜β‰ 0, since this new function 𝑔(π‘₯) is just a scaling of the original function 𝑓(π‘₯).

If the solution set consists of one element 𝑆={𝑠}, then the only root will be common with the vertex of the function. In this case, the function 𝑓(π‘₯)=(π‘₯βˆ’π‘ ) will have the given solution set, since 𝑓(𝑠)=0 by definition. As with the previous situation, we can multiply the original function 𝑓(π‘₯) by a nonzero constant, meaning that the new function 𝑔(π‘₯)=π‘˜π‘“(π‘₯) will have the same solution set.

We can actually make a more inclusive statement regarding quadratic functions that share the solution set βˆ…. For a general quadratic function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, where π‘Ž, 𝑏, and 𝑐 are real numbers and π‘Žβ‰ 0, we argued earlier using the quadratic root formula that there would be not real roots of 𝑓(π‘₯) if 𝑏<4π‘Žπ‘οŠ¨. This is because we would be attempting to take the square root of a negative number, which is not allowed for real numbers. In this situation, the solution set is always the empty set βˆ…. Therefore, we can state that any quadratic where 𝑏<4π‘Žπ‘οŠ¨ will share a solution set.

In the following examples, we will attempt to answer the questions in a fairly brisk manner, but we will at points reiterate this point above by demonstrating a range of quadratic functions that hold the desired properties.

Example 5: Understanding Zeros of Quadratic Functions

If the graph of the quadratic function 𝑓 cuts the π‘₯-axis at the points (βˆ’3,0) and (βˆ’9,0), what is the solution set of 𝑓(π‘₯)=0 in ℝ?

Answer

There are two points where the graph cuts the π‘₯-axis: at π‘₯=βˆ’3 and π‘₯=βˆ’9. These are called the roots of the function and this implies that the solution set is written as 𝑆={βˆ’3,βˆ’9}. This solution set will be common to any quadratic of the form 𝑓(π‘₯)=π‘˜(π‘₯+3)(π‘₯+9), where π‘˜β‰ 0, since 𝑓(π‘₯) will have a solution when π‘₯=βˆ’3 and π‘₯=βˆ’9.

Example 6: Finding the Solution Set of a Quadratic Equation given the Vertex Point of the Related Quadratic Function

If the point (9,0) is the vertex of the graph of the quadratic function 𝑓, what is the solution set of the equation 𝑓(π‘₯)=0?

Answer

Here, the case is that the vertex of 𝑓(π‘₯) is also the only root, since it occurs at the point (9,0). This means that the solution set only has one element and is written as follows: 𝑆={9}. This solution set will be common to any quadratic function of the form 𝑓(π‘₯)=π‘˜(π‘₯βˆ’9), where π‘˜β‰ 0.

Example 7: Understanding the Geometry of a Quadratic Function and Using This to Determine Its Solution Set

True or False: If the point (βˆ’2,βˆ’5) is the vertex of the graph of the quadratic function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, where π‘Ž is a negative number, then the solution set of the equation 𝑓(π‘₯)=0 is βˆ….

Answer

Given that π‘Ž<0, the quadratic function will have an β€œn” shape as opposed to the β€œu” shape that is found when π‘Ž>0. This means that the vertex is the maximum point of the function 𝑓(π‘₯), so the function cannot be above this point. Alternatively, we could say that 𝑓(π‘₯)β‰€βˆ’5, hence being bound above by the 𝑦-coordinate of the maximum point. Given that 𝑓(π‘₯)β‰€βˆ’5 for all values of π‘₯, it can never be the case that 𝑓(π‘₯)=0, meaning that there are no roots of the function. This means that the solution set 𝑆 is empty, and this is being represented as 𝑆=βˆ….

Key Points

  • A quadratic can be graphed using a table of values and this can be used to approximate the points at which the graph crosses the π‘₯-axis. These points are known as the β€œroots” of the quadratic.
  • Consider the quadratic function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, where π‘Ž, 𝑏, and 𝑐 are real numbers and π‘Žβ‰ 0. Then, the solution set 𝑆 of the function 𝑓(π‘₯) is the set consisting of values of π‘₯ such that 𝑓(π‘₯)=0. In other words,
    • 𝑏>4π‘Žπ‘οŠ¨; then 𝑆=𝑠=βˆ’π‘+βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž,𝑠=βˆ’π‘βˆ’βˆšπ‘βˆ’4π‘Žπ‘2π‘ŽοΌοŠ§οŠ¨οŠ¨οŠ¨ and the function can be written as 𝑓(π‘₯)=π‘˜(π‘₯βˆ’π‘ )(π‘₯βˆ’π‘ ), where π‘˜β‰ 0;
    • 𝑏=4π‘Žπ‘οŠ¨; then 𝑆=𝑠=βˆ’π‘2π‘ŽοΉοŠ§ and the function can be written as 𝑓(π‘₯)=π‘˜(π‘₯βˆ’π‘ ), where π‘˜β‰ 0;
    • 𝑏<4π‘Žπ‘οŠ¨; then 𝑆=βˆ….
  • When the solution set only has one element, the graph of 𝑓(π‘₯) crosses the π‘₯-axis exactly once, at the vertex (either a minimum or maximum point).

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