What if you and a friend started running from the same spot. You ran west and your friend ran south. After two hours, you had run 10 miles. The distance at that point between you and your friend was three times the distance your friend ran plus 2 miles? How could you determine how many miles your friend ran? After completing this Concept, you'll be able to solve real-world quadratic equation problems like this one.
CK-12 Foundation: 1009S Solving Real-World Problems Using Quadratic Equations
In mathematics, you’ll need to solve quadratic equations that describe application problems or that are part of more complicated problems. You’ve learned four ways of solving a quadratic equation:
- Taking the square root
- Quadratic formula
- Completing the square
Usually you’ll have to decide for yourself which method to use. However, here are some guidelines as to which methods are better in different situations.
Factoring is always best if the quadratic expression is easily factorable. It is always worthwhile to check if you can factor because this is the fastest method. Many expressions are not factorable so this method is not used very often in practice.
Taking the square root is best used when there is no term in the equation.
Quadratic formula is the method that is used most often for solving a quadratic equation. When solving directly by taking square root and factoring does not work, this is the method that most people prefer to use.
Completing the square can be used to solve any quadratic equation. This is usually not any better than using the quadratic formula (in terms of difficult computations), but it is very useful if you need to rewrite a quadratic function in vertex form. It’s also used to rewrite the equations of circles, ellipses and hyperbolas in standard form (something you’ll do in algebra II, trigonometry, physics, calculus, and beyond).
If you are using factoring or the quadratic formula, make sure that the equation is in standard form.
Solve each quadratic equation.
a) This expression if easily factorable so we can factor and apply the zero-product property:
b) Since the expression is missing the term we can take the square root:
Take the square root of both sides: and
c) Re-write the equation in standard form:
It is not apparent right away if the expression is factorable so we will use the quadratic formula:
Answer: no real solution
d) This problem can be solved easily either with factoring or taking the square root. Let’s take the square root in this case:
Solving Real-World Problems by Completing the Square
In the last section you learned that an object that is dropped falls under the influence of gravity. The equation for its height with respect to time is given by , where represents the initial height of the object and is the coefficient of gravity on earth, which equals or .
On the other hand, if an object is thrown straight up or straight down in the air, it has an initial vertical velocity. This term is usually represented by the notation . Its value is positive if the object is thrown up in the air and is negative if the object is thrown down. The equation for the height of the object in this case is
Plugging in the appropriate value for turns this equation into
if you wish to have the height in meters
if you wish to have the height in feet
An arrow is shot straight up from a height of 2 meters with a velocity of 50 m/s.
a) How high will the arrow be 4 seconds after being shot? After 8 seconds?
b) At what time will the arrow hit the ground again?
c) What is the maximum height that the arrow will reach and at what time will that happen?
Since we are given the velocity in m/s, use:
We know and meters so:
a) To find how high the arrow will be 4 seconds after being shot we plug in :
we plug in :
b) The height of the arrow on the ground is , so:
Solve for by completing the square:
The arrow will hit the ground about 10.2 seconds after it is shot.
c) If we graph the height of the arrow with respect to time we would get an upside down parabola . The maximum height and the time when this occurs is really the vertex of this parabola: .
The vertex is at (5.1, 129.45). In other words, when , the height is .
Another type of application problem that can be solved using quadratic equations is one where two objects are moving away from each other in perpendicular directions. Here is an example of this type of problem.
Two cars leave an intersection. One car travels north; the other travels east. When the car traveling north had gone 30 miles, the distance between the cars was 10 miles more than twice the distance traveled by the car heading east. Find the distance between the cars at that time.
Let the distance traveled by the car heading east
Then the distance between the two cars
Let’s make a sketch:
We can use the Pythagorean Theorem to find an equation for :
Expand parentheses and simplify:
Solve by completing the square:
Since only positive distances make sense here, the distance between the two cars is:
Solve Applications Using Quadratic Functions by any Method
Here is an application problem that arises from number relationships and geometry applications.
The product of two positive consecutive integers is 156. Find the integers.
Define: Let the smaller integer
Then the next integer
Translate: The product of the two numbers is 156. We can write the equation:
Apply the quadratic formula with:
Since we are looking for positive integers, we want . So the numbers are 12 and 13.
Check: . The answer checks out.
Suzie wants to build a garden that has three separate rectangular sections. She wants to fence around the whole garden and between each section as shown. The plot is twice as long as it is wide and the total area is . How much fencing does Suzie need?
Define: Let the width of the plot
Then the length of the plot
Translate: area of a rectangle is , so
Solve by taking the square root:
We take since only positive dimensions make sense.
The plot of land is .
To fence the garden the way Suzie wants, we need 2 lengths and 4 widths feet of fence.
Check: and . The answer checks out.
Watch this video for help with the Examples above.
CK-12 Foundation: 1009 Solving Real-World Problems Using Quadratic Equations
- For a quadratic equation in standard form, , the quadratic formula looks like this:
- A perfect square trinomial has the form , which factors into .
1. An isosceles triangle is enclosed in a square so that its base coincides with one of the sides of the square and the tip of the triangle touches the opposite side of the square. If the area of the triangle is what is the length of one side of the square?
2. The length of a rectangular pool is 10 meters more than its width. The area of the pool is 875 square meters. Find the dimensions of the pool.
1. Draw a sketch:
Define: Let base of the triangle
Then height of the triangle
Translate: Area of a triangle is , so
Solve by taking the square root:
The side of the square is 6.32 inches. That means the area of the square is , twice as big as the area of the triangle.
Check: It makes sense that the area of the square will be twice that of the triangle. If you look at the figure you can see that you could fit two triangles inside the square.
2. Draw a sketch:
Define: Let the width of the pool
Then the length of the pool
Translate: The area of a rectangle is , so we have .
Apply the quadratic formula with and
Since the dimensions of the pool should be positive, we want . So the pool is .
Check: . The answer checks out.
Solve the following quadratic equations using the method of your choice.
- The product of two consecutive integers is 72. Find the two numbers.
- The product of two consecutive odd integers is 1 less than 3 times their sum. Find the integers.
- The length of a rectangle exceeds its width by 3 inches. The area of the rectangle is 70 square inches, find its dimensions.
- Angel wants to cut off a square piece from the corner of a rectangular piece of plywood. The larger piece of wood is and the cut off part is of the total area of the plywood sheet. What is the length of the side of the square?
- Mike wants to fence three sides of a rectangular patio that is adjacent the back of his house. The area of the patio is and the length is 4 feet longer than the width. Find how much fencing Mike will need.
- Sam throws an egg straight down from a height of 25 feet. The initial velocity of the egg is 16 ft/sec. How long does it take the egg to reach the ground?
- Amanda and Dolvin leave their house at the same time. Amanda walks south and Dolvin bikes east. Half an hour later they are 5.5 miles away from each other and Dolvin has covered three miles more than the distance that Amanda covered. How far did Amanda walk and how far did Dolvin bike?