Lesson Objectives
  • Demonstrate an understanding of how to solve word problems
  • Learn how to set up and solve word problems with quadratic equations

How to Solve Word Problems with Quadratic Equations


We previously learned how to solve a word problem that involves setting up and solving a linear equation. Here, we will use the same six-step process with problems that involve quadratic equations.

Six-step method for Applications of Linear Equations in One Variable

  1. Read the problem carefully and determine what you are asked to find
    • Write down the main objective of the problem
  2. Assign a variable to represent the unknown
    • If more than one unknown exists, we express the other unknowns in terms of this variable
  3. Write out an equation that describes the given situation
  4. Solve the equation
  5. State the answer using a nice clear sentence
  6. Check the result by reading back through the problem
    • We need to make sure the answer is reasonable. In other words, if asked how many miles were driven to the store, the answer shouldn't be (-3) as we can't drive a negative amount of miles.
Example 1: Solve each word problem.
The area of a rectangular window is 143 square feet. If the length is 2 feet more than the width, what are the dimensions?
Step 1) Read the problem carefully and determine what you are asked to find:
Here, we are asked to find the dimensions of the window.
Step 2) Assign a variable to represent the unknown:
Here, we don't know the length or the width. We do know that the length is 2 feet more than the width.
let x = width of the window
then x + 2 = length of the window
Step 3) Write out an equation that describes the given situation:
To do this, let's make a little picture, which can help us to visualize the situation: picture of a rectangular window Area of a rectangle: length • width
We can multiply x, our width by (x + 2), our length, and set this equal to 143. $$x(x+2)=143$$ Step 4) Solve the equation: $$x(x+2)=143$$ $$x^2 + 2x=143$$ $$x^2 + 2x - 143=0$$ $$(x - 11)(x + 13)=0$$ $$x=11, -13$$ Since x represents the width of a window, we can throw out -13 since we can't have a width of -13 feet. This tells us that our only valid solution is 11 for the width and 13 (11 + 2) for the length.
Step 5) State the answer using a nice clear sentence:
The width of the window is 11 feet and the length is 13 feet.
Step 6) Check the result by reading back through the problem:
Reading back through the problem, we can see that our window has an area of 143 square feet. $$11 \text{ft}\cdot 13 \text{ft}=143 \text{ft}^2 ✓$$Additionally, we are told that the length is 2 feet more than the width. The length is 13 feet, which is 2 feet more than the width.
Example 2: Solve each word problem.
Two cars left an intersection at the same time. One of the cars traveled directly north, while the other car traveled directly east. Three hours later, they ended up being exactly 175 miles apart from each other. If the car that traveled north traveled 35 miles more than the car that traveled east, how far did each car travel?
Step 1) Read the problem carefully and determine what you are asked to find:
Here, we are asked to find how far each car traveled.
Step 2) Assign a variable to represent the unknown:
Here, we don't know how far the northbound car drove or how far the eastbound car drove.
We are told that the northbound car drove 35 miles more than the eastbound car.
let x = distance traveled by the eastbound car
then x + 35 = distance traveled by the northbound car
Step 3) Write out an equation that describes the given situation:
To do this, let's make a little picture, which can help us to visualize the situation: picture of the two cars forming a triangle and using the Pythagorean theorem From our image above, we can see that the Pythagorean formula applies to our problem. $$a^2 + b^2=c^2$$ $$x^2 + (x + 35)^2=175^2$$ Step 4) Solve the equation: $$x^2 + (x + 35)^2=175^2$$ $$x^2 + x^2 + 70x + 1225=30{,}625$$ $$2x^2 + 70x - 29{,}400=0$$ Use the quadratic formula: $$a=2, b=70, c=-29{,}400$$ $$x=\frac{-70 \pm \sqrt{70^2 - 4(2)(-29{,}400)}}{2(2)}$$ $$x=\frac{-70 \pm \sqrt{240{,}100}}{4}$$ $$x=\frac{-70 \pm 490}{4}$$ $$x=\frac{-70 + 490}{4}=\frac{420}{4}=105$$ $$x=\frac{-70 - 490}{4}=-\frac{560}{4}=-140$$ Since x represents the distance traveled by the eastbound car, we can throw out the -140. Recall that we can't have a negative amount for distance traveled. (Driving backwards does not count!)
This tells us that our only valid solution is 105 miles traveled by the eastbound car and 140 miles (105 + 35 = 140) traveled by the northbound car.
Step 5) State the answer using a nice clear sentence:
The eastbound car traveled a total distance of 105 miles, while the northbound car traveled a total distance of 140 miles.
Step 6) Check the result by reading back through the problem:
Reading back through the problem, we can see that our two cars are 175 miles apart at the three-hour mark. Additionally, based on the position of each, we can check using the Pythagorean formula. $$\sqrt{105^2 + 140^2}\stackrel{?}{=}175$$ $$175=175 ✓$$

Skills Check:

Example #1

Solve each word problem.

Olivia needs to wash a window in a building that is 16 feet from the ground. To avoid a fence, she decides to rest the ladder against the building. For stability, Olivia decides she should place the ladder 12 feet away from the building. How long of a ladder will Olivia need?

Please choose the best answer.

A
20 ft
B
22 ft
C
18 ft
D
29 ft
E
32 ft

Example #2

Solve each word problem.

Two cars left an intersection at the same time. One of the cars traveled directly north, while the other car traveled directly west into heavy traffic. An hour later, they ended up being exactly 41 miles apart from each other. If the car that traveled north traveled 31 miles more than the car that traveled west, how far did each car travel?

Note: NB -> northbound, WB -> westbound

Please choose the best answer.

A
WB: 15 mi, NB: 13mi
B
WB: 9 mi, NB: 40mi
C
WB: 27 mi, NB: 40mi
D
WB: 6 mi, NB: 37mi
E
WB: 8 mi, NB: 39mi

Example #3

Solve each word problem.

The length of a garden is 12 feet more than the width. If the area of the garden is 325 square feet, what are the dimensions of the garden? Note: (Area is length times width)

Please choose the best answer.

A
L: 22, W: 10
B
L: 25, W: 13
C
L: 27, W: 15
D
L: 30, W: 18
E
L: 33, W: 21
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