The portal has been deactivated. Please contact your portal admin.

Lesson Explainer: Applications of Quadratic Equations Mathematics

In this explainer, we will learn how to solve word problems by forming and solving quadratic equations.

Quadratic equations appear all throughout mathematics and in many real-world situations. For example, the height of a projectile traveling through air and the areas of rectangles can be modeled by quadratic equations.

We can apply all of the techniques we have for solving and analyzing the properties of quadratics to investigate and solve these real-world problems. For example, if we are told that the height β„Ž metres of a ball above the ground 𝑑 seconds after it is thrown is given by the quadratic equation β„Ž=βˆ’3𝑑+5𝑑+2, then we can model the path traveled by the ball by sketching the graph of this quadratic equation.

To help us sketch the graph, we can start by factoring the equation. We want two numbers whose product is βˆ’3Γ—2=βˆ’6 and whose sum is 5. We find that these are 6 and βˆ’1. We can factor by grouping by using these values to split the 𝑑-term: βˆ’3𝑑+5𝑑+2=βˆ’3𝑑+6π‘‘βˆ’π‘‘+2.

We take out the shared factor of βˆ’3𝑑 in the first two terms and the factor of βˆ’1 from the final two terms to get βˆ’3𝑑+6π‘‘βˆ’π‘‘+2=βˆ’3𝑑(π‘‘βˆ’2)βˆ’(π‘‘βˆ’2).

We can now take out the shared factor of π‘‘βˆ’2 to obtain β„Ž=(π‘‘βˆ’2)(βˆ’3π‘‘βˆ’1).

This gives us the horizontal intercepts of 𝑑=2 and 𝑑=βˆ’13. We can find the vertical intercept by substituting 𝑑=0 into the quadratic: β„Ž=βˆ’3(0)+5(0)+2=2.

We can sketch the path of the ball by sketching a parabola that opens downward with these intercepts.

Of course, we always need to consider the actual situation we are modeling before we conclude that our sketch is accurate and take information away from the diagram. We are modeling a ball flying through the air 𝑑 seconds after being thrown; this means that our model will not make sense when 𝑑 is negative. Similarly, we can say that when β„Ž=0, the ball will hit the ground, so we cannot use this quadratic model beyond this point. We can restrict our sketch to these values to obtain the following.

We can mark points on the parabola based on the real-world interpretation of the model. For example, when 𝑑=0, the value of β„Ž is the height of the ball in metres when the ball is initially thrown. We can see in the diagram that this is 2 m or, alternatively, it is the constant value in the quadratic equation β„Ž=βˆ’3𝑑+5𝑑+2.

Similarly, we note that when β„Ž=0, the ball hits the ground. We can see that this is when 𝑑=2 in the diagram, or, alternatively, we can find this using the factored form of the equation, noting that 𝑑=βˆ’13 is not a valid solution for the model since we start at 𝑑=0.

Finally, we know that the vertex of a parabola tells us the maximum (or minimum) output of the quadratic function. In this case, we have a maximum output at the vertex, so this will be the maximum height of the ball. We can find the coordinates of the vertex by completing the square. We first take out the factor of βˆ’3: βˆ’3𝑑+5𝑑+2=βˆ’3ο€Όπ‘‘βˆ’53π‘‘βˆ’23.

We then use the fact that 𝑑+𝑏𝑑+𝑐=𝑑+𝑏2ο‰βˆ’ο€½π‘2+π‘οŠ¨οŠ¨οŠ¨ to get βˆ’3ο€Όπ‘‘βˆ’53π‘‘βˆ’23=βˆ’3ο€Ύο€Όπ‘‘βˆ’56οˆβˆ’ο€Ό56οˆβˆ’23=βˆ’3ο€Όπ‘‘βˆ’56+4912.

Since the term βˆ’3ο€Όπ‘‘βˆ’56 is never positive, the expression will reach a maximum when this term is zero.

Hence, when 𝑑=56, we have a maximum output of β„Ž=4912. This is the maximum height that the ball is off the ground in metres.

In our first example, we will determine some information about a suspension bridge given a quadratic equation for its height from the ground.

Example 1: Analyzing a Quadratic Equation for the Height of a Suspension Bridge

The diagram shows part of a suspension bridge fastened onto two supports at the same height off the ground on both sides.

The height, β„Ž metres, of the suspension bridge from the ground is modeled by the equation β„Ž=25π‘‘βˆ’50𝑑+150, where 𝑑 is the horizontal distance from the left support in metres.

  1. Find the maximum height, in metres, of the suspension bridge from the ground.
  2. Find the minimum height, in metres, of the suspension bridge from the ground.

Answer

Part 1

The maximum height of the suspension will be the height of the suspension where it is fixed to the support. We can determine this by substituting 𝑑=0 into the given equation. We obtain β„Ž=25(0)βˆ’50(0)+150=150.m

Hence, the maximum height of the suspension bridge from the ground is 150 m.

Part 2

The minimum height of the suspension from the ground will occur at the vertex of the parabola. We recall that we can find the coordinates of the vertex of a parabola by completing the square. We first take out a factor of 25 from the right-hand side of the equation to get β„Ž=25π‘‘βˆ’50𝑑+150=25ο€Ήπ‘‘βˆ’2𝑑+6.

We can then use the fact that π‘₯+π‘Žπ‘₯+𝑏=ο€»π‘₯+π‘Ž2ο‡βˆ’ο€»π‘Ž2+π‘οŠ¨οŠ¨οŠ¨ to rewrite this equation as 25ο€Ήπ‘‘βˆ’2𝑑+6=25ο€Ί(π‘‘βˆ’1)βˆ’1+6=25(π‘‘βˆ’1)+125.

We know that 25(π‘‘βˆ’1)β‰₯0, for any value of 𝑑, so the lowest value of β„Ž will occur when this is 0. This occurs when 𝑑=1, so the minimum height occurs when 𝑑=1: β„Ž=25(1βˆ’1)+125=125.m

Hence, the minimum height of the suspension bridge from the ground is 125 m.

In our next example, we will determine the height of a building and the time taken for an object to hit the ground when dropped from the top of the building using a given quadratic equation.

Example 2: Analyzing a Quadratic Equation for Free Fall

A mass is dropped off the Leaning Tower of Pisa. The height, β„Ž metres, of the mass after 𝑑 seconds is modeled by the equation β„Ž=57βˆ’4.91π‘‘οŠ¨.

  1. Find the height of the Leaning Tower of Pisa.
  2. Find the time taken for the mass to hit the ground. Give your answer to two decimal places.

Answer

Part 1

In order to determine the height of the Leaning Tower of Pisa, we first note that we are given an equation for the height of the mass 𝑑 seconds after being dropped. If we set 𝑑=0, then we will have the initial height of the mass, which is the height of the building: β„Ž=57βˆ’4.91𝑑=57βˆ’4.91(0)=57.m

Hence, the height of the Leaning Tower of Pisa is 57 m.

Part 2

The equation we are given measures the height of the mass from the ground 𝑑 seconds after being dropped. This means that the mass will hit the ground when β„Ž=0. Substituting β„Ž=0 into the equation gives us 0=57βˆ’4.91𝑑.

We can solve this equation for 𝑑 by adding 4.91π‘‘οŠ¨ to both sides of the equation to obtain 4.91𝑑=57.

We then divide through by 4.91 to get 𝑑=574.91.

We now take square roots of both sides of the equation. This will give us a positive and a negative solution. However, our model starts at 𝑑=0, so we discard the negative root and only consider the positive root: 𝑑=ο„ž574.91=3.407….s

Rounding this value to two decimal places gives us 𝑑=3.41s.

In our next example, we will analyze a situation involving the height of a tunnel.

Example 3: Finding the Dimensions of a Parabolic Arch

The diagram shows an entrance to a tunnel with a parabolic arch connecting two straight vertical sections of a wall of equal height.

The height β„Ž of the tunnel in metres at 𝑀 metres from the left wall is given by the equation β„Ž=βˆ’12𝑀(π‘€βˆ’3)+2.

  1. Find the height of the straight vertical section of the wall.
  2. Find the total width of the arch.
  3. Find the maximum height of the arch.

Answer

Part 1

To find the height of the straight vertical sections of the wall, we note that the vertical sections of the wall are situated at the ends of the parabolic arch. This means that the height of these sections will be given by the value of β„Ž when 𝑀=0.

Substituting 𝑀=0 into the given equation yields β„Ž=βˆ’12(0)(0βˆ’3)+2=2.

Hence, the height of the straight vertical sections of the wall is 2 m.

Part 2

The total width of the arch will be given by the other value of 𝑀 for which the height is 2.

Substituting β„Ž=2 into the equation gives us 2=βˆ’12𝑀(π‘€βˆ’3)+2.

Subtracting 2 from the sides of the equation yields 0=βˆ’12𝑀(π‘€βˆ’3).

We can set each factor equal to 0 to see that there are two solutions, 𝑀=0 and 𝑀=3. Since 𝑀=0 is the start of the arch, we must have that the total width of the arch is 3 m.

Part 3

In order to find the maximum height of the arch, we first note that the maximum height of the arch is the maximum output of the quadratic. We then recall that the maximum output of a quadratic function occurs at its vertex and that we can find this output by completing the square. We first expand the brackets to get β„Ž=βˆ’12𝑀(π‘€βˆ’3)+2=βˆ’12𝑀+32𝑀+2.

We then take out a factor of βˆ’12 to obtain βˆ’12𝑀+32𝑀+2=βˆ’12ο€Ήπ‘€βˆ’3π‘€βˆ’4.

We can now complete the square using the fact that π‘₯+π‘Žπ‘₯+𝑏=ο€»π‘₯+π‘Ž2ο‡βˆ’ο€»π‘Ž2+π‘οŠ¨οŠ¨οŠ¨. This gives us βˆ’12ο€Ήπ‘€βˆ’3π‘€βˆ’4=βˆ’12ο€Ύο€Όπ‘€βˆ’32οˆβˆ’ο€Ό32οˆβˆ’4=βˆ’12ο€Όπ‘€βˆ’32+258.

We see that the maximum output of the function occurs when 𝑀=32. Therefore, the maximum height of the tunnel is 258 m.

In our next example, we will use a given quadratic equation for the value of a vintage car to determine the best time to buy the car and how long it would take for the car to regain its value.

Example 4: Analyzing the Price of a Vintage Car Using Quadratics

The value 𝑉 of a vintage car in pounds 𝑑 years after the yearΒ 1980 is given by the equation 𝑉=15π‘‘βˆ’20𝑑+5000.

  1. What is the value of the car in 1980?
  2. How many years after 1980 does it take for the car to reach its original value?
  3. How many years after 1980 does the value of the car reach its lowest value?

Answer

Part 1

Since the value of the car 𝑑 years after the yearΒ  1980 is given by the equation 𝑉=15π‘‘βˆ’20𝑑+5000, we can determine the value of the car in various years by substituting values of 𝑑 into the equation. If we substitute 𝑑=0 into the equation, we will find the value of the car 0 years after 1980, which is 1980. However, we do not need to do this directly; we can also recall that the initial value of a quadratic is equal to the constant in the polynomial, in this case Β£5β€Žβ€‰β€Ž000.

Hence, the value of the car in 1980 is Β£5β€Žβ€‰β€Ž000.

Part 2

We found the original value of the car in part 1 to be Β£5β€Žβ€‰β€Ž000. For the car to regain its value, we must have 𝑉=5000. Substituting this into the equation gives 5000=15π‘‘βˆ’20𝑑+5000.

Subtracting 5β€Žβ€‰β€Ž000 from both sides of the equation yields 0=15π‘‘βˆ’20𝑑.

We can take out a factor of 15𝑑 to obtain 0=15𝑑(π‘‘βˆ’100).

We can solve this equation by setting each factor equal to 0 and solving. We have either 𝑑=0 or 𝑑=100. We know that the solution 𝑑=0 corresponds to the initial value, so we disregard this solution.

Hence, the car will regain its value 100 years after 1980.

Part 3

The lowest value of the car will occur at the lowest output of the quadratic function 15π‘‘βˆ’20𝑑+5000. Since this is a quadratic with a positive leading coefficient, we know it will open upward, and so its vertex will be the minimum of the function. Therefore, we can find the lowest value of the car and the time this occurs by finding the coordinates of the vertex.

We recall that to find the coordinates of the vertex, we want to complete the square. We first take out the factor of 15 to get 𝑉=15π‘‘βˆ’20𝑑+5000=15ο€Ήπ‘‘βˆ’100𝑑+25000.

We can now complete the square by recalling that π‘₯+π‘Žπ‘₯+𝑏=ο€»π‘₯+π‘Ž2ο‡βˆ’ο€»π‘Ž2+π‘οŠ¨οŠ¨οŠ¨. Thus, 15ο€Ήπ‘‘βˆ’100𝑑+25000=15ο€Ί(π‘‘βˆ’50)βˆ’50+25000=15(π‘‘βˆ’50)+4500.

Since the term 15(π‘‘βˆ’50) is nonnegative for any value of 𝑑, we can conclude that the minimum value occurs when this term is 0. That is, when 𝑑=50, the minimum value will be 𝑉=4500Β£.

Hence, the minimum value of the car occurs 50 years after 1980.

In our final example, we will use the context of a real-world problem and some given measurements to determine unknown values in a quadratic equation.

Example 5: Finding Unknowns in a Quadratic Using Context and Measurements

A ball is thrown from a building and its height β„Ž in metresΒ  𝑑 seconds after being thrown follows the quadratic equation β„Ž=βˆ’π‘‘+𝑏𝑑+π‘οŠ¨. The building is 40 m tall. The maximum height of the ball off the ground is 44 m and this height is reached after 2 s.

Find the values of 𝑏 and 𝑐.

Answer

The initial height of the ball is found by substituting 𝑑=0 into the equation; this will give the height of the building. Thus, we substitute β„Ž=40 and 𝑑=0 into the equation to get 40=βˆ’(0)+𝑏(0)+𝑐=𝑐.

Hence, 𝑐=40.

Next, we are told that the maximum height of the ball off the ground is 44 m and that this height is reached after 2 s. We know that this occurs at the vertex of the parabola, and we recall that we can find the coordinates of the vertex by completing the square: β„Ž=βˆ’π‘‘+𝑏𝑑+40=βˆ’ο€Ήπ‘‘βˆ’π‘π‘‘βˆ’40=βˆ’ο€Ώο€½π‘‘βˆ’π‘2ο‰βˆ’π‘4βˆ’40=βˆ’ο€½π‘‘βˆ’π‘2+𝑏4+40.

We know that the vertex will occur when the term βˆ’ο€½π‘‘βˆ’π‘2ο‰οŠ¨ is equal to 0. Thus, the maximum height of the ball occurs when π‘‘βˆ’π‘2=0. We are told that this happens when 𝑑=2s, so 2βˆ’π‘2=0𝑏=4.

Hence, 𝑏=4 and 𝑐=40.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • Many real-world problems involve quadratic equations, and we can apply all of the techniques we have for solving and analyzing the properties of quadratics to investigate and solve these real-world problems.
  • We can factor quadratics or use the quadratic formula to find the roots of a quadratic; we can find the 𝑦-intercept by substituting π‘₯=0 into the equation or by finding the constant in the equation.
  • We can find the maximum and minimum values using the coordinates of the vertex, which we get by completing the square.
  • In real-world problems, we always need to consider the validity of our solutions. For example, we may discount negative solutions since they may not be possible for the real-world property we are measuring.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.