In this explainer, we will learn how to solve word problems by forming and solving quadratic equations.
On occasion, we will be tasked with forming and solving quadratic equations based on a real-world scenario. Quite often, these will be problems involving area and finding unknown lengths or any other field in which quadratic equations appear.
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: Finding the Perimeter of a Rectangle given Its Area and the Difference between Its Dimensions
What is the perimeter of a rectangle whose length is 7 cm more than its width and whose area is 78 cm2?
Answer
To start, it can be helpful to draw a diagram representing the scenario described. We know that the length is 7 cm more than the width, so let us call the width , in centimetres, and the length . This gives us the following rectangle.
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 . Since the area is 78, we can use this to form the following equation:
If we then use the distributive property to expand the brackets, we get
Subtracting 78 from both sides gives us
At this point, we have a quadratic in a form that can be solved. We can check whether the equation can be factored, or we can solve by completing the square or using the quadratic formula. If we consider the factor pairs of 78, we have
We need two numbers that multiply to make and add to make 7; using the factor pairs, we see these are and 13.
Therefore, the equation factors as follows:
Our solutions for are then calculated by identifying the points at which each of the two factors are equal to zero. That is, and .
Given that is a length, it cannot be negative, so our solution must be .
Finally, we need to determine the perimeter of the rectangle—that is, the sum of the side lengths. We know the width is 6 cm and the length is , so the perimeter is given by
In our previous example, we noticed a key difference between solving quadratic equations in a mathematical sense and in a real-world context. We saw that the underlying quadratic had two solutions; however, since was a length, it had to be positive, meaning one of the solutions was not valid. It is always a good idea to check the answers found at the end of these types of problems to make sure they make sense in the context of their real-world scenario.
We will see a similar scenario in our next question involving finding a positive number that satisfies a certain property.
Example 2: Forming and Solving Quadratic Equations
Determine the positive number whose square exceeds twice its value by 15.
Answer
The first thing we need to do is translate the wording of the problem into an equation. 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, we can observe that . Hence, we are able to factor the equation as follows:
Thus, either or . Solving each equation we get and . However, we were told that we are looking for a positive number; therefore, the solution is 5.
In our next example, we will solve another problem involving lengths in a rectangle, this time by using the quadratic formula.
Example 3: Using Quadratic Equations to Solve Problems
A right triangle has sides of lengths cm, cm, and cm. Find the length of its shortest side.
Hint: First, decide which side is the hypotenuse.
Answer
Since this is a problem involving lengths in a right triangle, it is a good idea to sketch the information we are given. To do this, we first need to determine which length is the hypotenuse. We note that one of the sides is of length cm, so must be positive. We also note that is smaller than the other length, , so this must be the hypotenuse. This gives us the following.
The shortest side has length , so we need to determine the value of . We can do this by applying the Pythagorean theorem to the right triangle; we get
Distributing the exponents over the parentheses gives us
We can then collect like terms and rearrange to form a quadratic equation:
We can solve this by finding a pair of numbers whose product is and whose sum is . We see that this is and 1, so we have
For a product to be equal to zero, one of the factors must be equal to zero. Hence, we either have or . Since is a side length, it must be positive, so .
Finally, the shortest side of the triangle has length cm, so the shortest side is of length 7 cm.
Before we continue, it is worth noting that all the of the unique problems we have with quadratics also occur when applying them to real-world scenarios. For example, we cannot always factor quadratics and will sometimes need to apply the quadratic formula to find the roots that we can then round to an appropriate degree of accuracy. Another possibility is that there could be 0,1, or 2 solutions depending on the sign of the discriminant and there are still the real-world limitations on the possible values of the variables.
Let us see an example of a problem in which both of these situations occur. Imagine that the entrance to a tunnel is modeled by a parabola with equation , where is the height of the tunnel in metres and is the horizontal displacement in metres from a known point, as shown.
We want to use this to find the maximum width of the tunnel entrance. We can first note that parabolas open outward, so the maximum width occurs at the lowest part of the tunnel, when . Therefore, we can find the values of these points by solving the equation
We cannot solve this by factoring, so we will use the quadratic formula, which states that the solutions to the quadratic are
In this quadratic, we have , , and . Substituting these values in and simplifying, we have
Since, the expression for the roots does not simplify any further, we will give the roots to two decimal places. We have and . The maximum width is then the difference between these values, which we find by using the exact values of the roots. To two decimal places, this is 3.20 m.
In our final three examples, we will solve various problems in geometry and number theory by forming and solving quadratic equations.
Example 4: Using Quadratic Equations to Solve Problems
The diagram shows a rectangular prism, where the area of its net is 580. Find the value of .
Answer
We first note that the area of the net of this rectangular prism will be the same as its surface area. In other words, the sum of the area of the faces of the rectangular prism is 580 square units. We need to start by finding an expression for this sum. To do this, we note that there are 6 faces and opposite faces have the same area. Finally, we know that each face is either a rectangle or a square, so the area of each face is length times width. This allows us to find the area of each face as follows.
The length of the front face is and its width is 3, so its area is . The length of the top face is and its width is , so its area is . The length of the side faces is and their width is 3, so their area is .
Since there are two of each face, the surface area is the sum of two times each expression, giving
We are told this is equal to 580, giving us the equation
We subtract 580 from both sides to get
We could then attempt to solve this by factoring; however, the product of has many factor pairs, so it is easier to apply the quadratic formula. We recall the quadratic formula states that the solutions to the quadratic are
In this quadratic, we have , , and . Substituting these values in and simplifying gives us
Evaluating each root separately, we get that or . Since represents a length, it must be positive; hence, .
Example 5: Forming and Solving Quadratic Equations
Find the positive number that is 66 less than twice its square.
Answer
If we call the positive number we are trying to find , then we are told that is 66 less than twice its square. Twice the square of is , and since is 66 less than this, we can add 66 to to create an equivalent expression, so we have
We can solve this quadratic equation for by first rearranging the equation to get
Next, we need to find two numbers that multiply to give and add to give . By considering the factor pairs of 132, we can see that these are and 11. We use these to rewrite the equation as
We now take out the shared factor of from the first two terms and the shared factor of 11 in the last two terms to get
Finally, we take out the shared factor of , giving us
For the product of two numbers to be zero, one of the factors must be zero. Hence, either or . Solving each equation, we get or . Since we are told is positive, we have .
Hence, the positive number that is 66 less than twice its square is 6.
In our final example, we will construct and solve a quadratic equation from a geometric problem that involves equating the areas of a trapezoid and a rectangle.
Example 6: Using Quadratic Equations to Solve Problems
The diagram shows a trapezoid and a rectangle.
- Write an expression for the area of the rectangle.
- Write an expression for the area of the trapezoid.
- If the trapezoid and the rectangle have the same area, find the value of using a suitable equation.
Answer
Part 1
We recall that the area of a rectangle is width times length. In the diagram, the rectangle has a length of and a width of . Hence, its area is the product of the expressions:
Part 2
We recall that the area of a trapezoid is half the sum of the parallel sides (or base sides) multiplied by the perpendicular height. We see in the picture that the parallel sides have lengths and , so their sum is . We can also note that the perpendicular height is . So, the area of this trapezoid is half the product of these expressions, which is given by
Part 3
If the trapezoid and rectangle have equal areas, then the expressions for their areas must be equal; equating these expressions, we have
Distributing the parentheses on the left-hand side of the equation, we have
Distributing the parentheses on the right-hand side of the equation, we have
Equating the expansions of each expression for the areas gives us
We can then rearrange this into a quadratic equation in standard form as follows:
Next, we can solve this equation using factoring by noting that and ; this gives
Then, we can equate each factor to zero to see that and are solutions to the equation.
We might be tempted to conclude that both of these values are solutions since they are both positive; however, we see that the rectangle has a length of . This means that cannot be smaller than 9; otherwise, this length would be negative.
Therefore, the value of is 12.
Let us finish by recapping some of the important points of this explainer.
Key Points
- In some problems, we may need to extract the information in the question and use this to form a quadratic equation.
- It is often necessary to simplify the quadratic equation into standard form so that it can be solved.
- We can use any of the standard methods to solve the quadratic, which include factoring, completing the square, factoring by grouping, and using the quadratic formula.
- We should check any solutions with the context of the problem; sometimes, only one solution will be appropriate in the given context.
