Solving Word problems using 1 variable in nonlinear equations by factoring

Click here for videos on Factoring.

Max and Min Problems--uses the idea of the vertex of a parabola to find a minimum or maximum value

WP stands for Word Problem...

WP1-Solve number puzzle by factoring

This is a Word Problem you solve by writing an equation and factoring. Word Problem 1 - Number Puzzle-The square of a negative number is 15 more than twice the negative number. What is the number?

http://www.youtube.com/watch?v=HOxLak15pY8

WP2-Solve Triangle if you know perimeter

This is a Word Problem you solve by writing an equation and factoring. Word Problem 2 - Perimeter of a Triangle-The perimeter of a given triangle is 85 meters. By looking at the picture, state the lengths of the sides of the triangle.

http://www.youtube.com/watch?v=AcGK5JL_aqk

WP3-Solve Rectangle if you know area

This is a Word Problem you solve by writing an equation and factoring. Word Problem 3 - Area of a Rectangle-Use the info from a picture of a rectangle to find the length and width is its area is 54 sq. ft.

http://www.youtube.com/watch?v=1DE17TuY8Rw

WP4-Solve triangle if you know area

This is a Word Problem you solve by writing an equation and factoring. Word Problem 4 -Area of a Triangle-The height of a triangle is 4 inches more than twice the length of the base. The area of the triangle is 35 square inches. Find the height of the triangle.

http://www.youtube.com/watch?v=EZjhKZU4Rko

WP5-Solve triangle using Pythagorean theorem

This is a Word Problem you solve by writing an equation and factoring. Word Problem 5 - Pythagorean Theorem-Find the lengths of the sides of the right triangle shown on the diagram.

http://www.youtube.com/watch?v=5TPrdcpgtX8

WP6-Find time using physics formula

This is a Word Problem you solve by writing an equation and factoring. Word Problem 6 - Physics ball falling application-Use the formula given to do this problem. A ball is thrown on a hard surface and bounces straight up. The initial velocity of the rebound is 64 feet per second. How many seconds later will the ball hit the ground?

http://www.youtube.com/watch?v=CepqshwePog

WP7-Find interest rate based on investment formula

This is a Word Problem you solve by writing an equation and factoring. Word Problem 7 - Investment problem-At the end of 2 years, P dollars invested at an interest rate r compounded annually increases to an amount, A dollars, given by a formula shown in video. Find the interest rate if $2000 increased to $2420 in 2 years. Write the answer as a percent.

http://www.youtube.com/watch?v=5vGMzZpXEBg

WP8-Solve triangle using Pythagorean theorem

This is a Word Problem you solve by writing an equation and factoring. Word Problem 5 - Pythagorean Theorem-Find the lengths of the sides of the right triangle shown on the diagram.

http://www.youtube.com/watch?v=zv7yStIhF-4

Below are two more examples. I have not created videos for these, but solutions are at bottom.

1. The length of a rectangle is 2 inches longer than three times the width. If the area is 56 square inches, what are the dimensions?

2. The perimeter of a rectangle is 36 feet. If the area is 56 square feet, what are the dimensions?

Solutions are below.

EXAMPLE 1

The length of a rectangle is 2 inches longer than three times the width. If the area is 56 square inches, what are the dimensions?

Width: w

Length:  3w+2

w(3w+2)=56

3w^2+2w=56

3w^2+2w-56=0

(3w+ 14)(w-4)=0   

3w+14=0 gives w=-14/3 which can't be the width since it is negative.

w-4=0 gives w=4 so the width is 4 and length is 3·4+2=14

Width: 4 in.

Length: 14 in.

Ck: The length is 2 inches more than three times the width, and the area is 56 square inches since 4·14=56

EXAMPLE 2:

The perimeter of a rectangle is 36 feet. If the area is 56 square feet, what are the dimensions?

Width: W

Length:  L

If the perimeter is 36, then half way around is 18, which is W+L. So W+L=18. That means L=18—W

So instead let's define our variables with one variable:

Width: W

Length: 18—W

Since the area is 56, we get

W(18—W)=56

18W—W^2=56

0=W^2—18W+56

0=(W—14)(W—4)

W = 14 or W = 4

Possibility 1: The width is 14, the length is 18—14=4 so the dimensions are 14 ft and 4 ft.

Possibility 2: The width is 4, the length is 18—4=14 so the dimensions are 4 ft and 14 ft.

Answer in either case: 4 ft and 14 ft.