In this explainer, we will learn how to solve word problems by forming and solving quadratic equations.
On occasion, you will be asked to solve story problems which involve forming and solving quadratic equations. Quite often, these will be problems involving area and finding unknown lengths. The standard approach with these questions is to extract the key information from the question and use this to form an equation that can then be solved using standard methods. To demonstrate this, we will look at a series of examples, looking carefully at each of the steps needed to reach a solution.
Example 1: Using Quadratic Equations to Solve Story Problems
The length of a rectangle is 2 cm more than its width. If its area is 80 cm2, what are its length and width?
To start, it can be helpful to draw a diagram representing the situation described. We know that the length is 2 cm more than the width, so let us call the width and the length .
We know that the area of a rectangle is calculated by multiplying its length by its width. Here, the length is and the width is , so we can use this to form the following equation:
If we then use the distributive property to expand the brackets, we get
Subtracting 80 from both sides gives us
At this point, we have a quadratic in a form that can be solved. We can check if the equation can be factorized, or we can solve by completing the square or using the quadratic formula. If we consider the factor pair of 80, we have
Of these, we can make 2 using , that is,
Therefore, the equation factors as follows:
Our solutions for are then calculated by identifying the points at which each of the two brackets is zero. That is, and . Given that is a length, it cannot be negative so our solution must be . The width is then 8 and the length is 10.
Example 2: Using Quadratic Equations to Solve Word Problems
The length of a rectangle is 26 cm more than its width. Given that its area is 120 cm2, determine its perimeter.
We are told that the length of the rectangle is 26 more than its width, so if we call the width , then the length must be . It can then be helpful to draw a diagram to help visualize the situation.
Remember that the area of a rectangle is calculated by multiplying its length by its width. Using this, we can form an equation:
Expanding the brackets, we get and subtracting 120 from each side gives us
At this point, we have a quadratic that can be solved. You may notice that this particular equation can be factored as a factor pair of 120 is , but alternatively we can solve this using the quadratic formula. For a quadratic in the form we have that
For this equation, we have , , and . Substituting, we get that and then calculating for both the plus and the minus gives values of 4 and . As is a length, it cannot be ; therefore, the width must be 4 and the length 30. A common mistake here would be to stop at this point, but the question asks for the perimeter of the rectangle. The formula for the perimeter of a rectangle is where is the length of the rectangle and is the width. Substituting into this, we get
Note that, in the previous two examples, the questions were very similar and the method of calculation was largely the same, although, in the second example, we opted to solve the quadratic equation using a different method. Also, the second example asked for an additional calculation which is very easily missed out if the question is not read carefully.
Let us continue by looking at some more similar, but subtly different, examples.
Example 3: Using Quadratic Equations to Solve Story Problems
The side length of a square is cm, and the dimensions of a rectangle are cm and 2 cm. Given that the sum of their areas is 8 cm2, determine the perimeter of the square.
We are told that we have a square of length , the area of which can be calculated by squaring its length, so the square has an area of . Similarly, we are told that we have a rectangle with dimensions of and 2. A rectangle’s area is calculated by multiplying its length by its width, so this has an area of . The sum of their areas, we are told, is 8, so we can form the equation
If we then subtract 8 from both sides, we have
This can then be factored to
By calculating when each of the two expressions in the parentheses is zero, we find that our solutions are and . As is a length, the solution cannot be negative, which means the solution is . The question then asks for the perimeter of the square which is four times the length of one of its sides, so, the perimeter is
Example 4: Forming Quadratic Equations Representing Word Problems
The length of a rectangle is 3 cm more than double the width. The area of the rectangle is 27 cm2. Write an equation that can be used to find , the width of the rectangle, in centimetres.
Here, we are told that the length is 3 more than double the width, . This tells us that the length is . We can calculate the area of the rectangle by multiplying the length by the width, that is,
We are told that the area is equal to 27, so we can equate our expression and 27 to form an equation:
If we then expanded the parentheses and subtracted 27 from both sides, we would have a quadratic in standard form which we could then solve.
Example 5: Using Quadratic Equations to Solve Word Problems
A rectangular photograph measuring 6 cm by 4 cm is to be displayed in a card mount in a rectangular frame, as shown in the diagram.
Which of the following equations can be used to find if the area of the mount is 64 cm2?
We start by considering the larger, outside, rectangle; this has a length , a width , and, therefore, an area
The smaller rectangle has an area of
If we then subtract the smaller rectangle from the larger rectangle, we get which is an expression for the area of the card mount. We are told that this is equal to 64 which allows us to form the equation
If we wanted to solve this, we would expand the parentheses, write the equation in standard from, and then solve this using standard methods to find .
Example 6: Forming and Solving Quadratic Equations
Determine the positive number whose square exceeds twice its value by 15.
The first thing we need to do is translate the wording of the problem into equations. We let be the number we are trying to find. The first piece of information we are given is that is positive; that is, . Secondly, we are told the square of the number, , exceeds twice its value, , by 15. This means that the difference between and is 15. Hence,
Subtracting 15 from both sides of the equation, we get the following quadratic in standard form:
We can now solve this equation to find the value of . Notice that 15 is the product of 3 and 5. Furthermore, notice that . Hence, we are able to factor the equation as follows:
Therefore, the solutions to this equation are and . However, we were told that we are looking for a positive number; therefore, the solution is 5.
Example 7: Solving Quadratic Equations in Real-World Contexts
A formula for the normal systolic blood pressure for a man aged , measured in millimeters of mercury, is given as . Find the age, to the nearest year, of a man whose normal systolic blood pressure measures 125 mmHg.
We are trying to find when . Therefore, we have
Subtracting 125 from both sides of the equation, we get the following quadratic in standard form:
To solve this equation, we can use the quadratic formula:
Substituting in the values , , and , we have
Using a calculator, we can evaluate this expression to get and . Since we are looking for the age of a person, we can discard the negative solution and conclude that to the nearest year.
- Read the question carefully to identify the important information that can be used to form a quadratic equation.
- Simplify the quadratic equation into standard form so that it can be solved.
- Use standard methods to solve the quadratic, which include factoring, completing the square, and using the quadratic formula.
- Check the two solutions with the context of the problem. Sometimes, only one solution will be appropriate in the given context.