Algebra Honors ( Period 5 & 6)

Using Factoring to Solve Problems  5-13

Example 1

A decorator plans to place a rug in a 8 m by 12 m room so that a uniform strip of wood flooring around the rug will remain uncovered. How wide will this strip be if the area of the rug is to be half the area of the room?

Let x = the width of the strip

Then 12-2x is the length of the rug and 9 – 2x is the width of the rug

(12-2x)(9-2x) = the area of the rug.

Area of the rug = ½ ( area of the room)

(12-2x)(9-2x) = ½(9∙12)

108 – 42x + 4x² = 54

4x²- 42 + 108 = 54

4x²- 42 -54 = 0

2(2x²- 21 + 27) =0

2(2x -3)(x -9) = 0

using the ZERO Products Property

2x -3 = 0 or x= 3/2

x -9 = 0  x = 9

CHECK: x = 1.5

works but when x = 9

the length 12 – 2x and the width 9-2x are negative! Since a negative length or width is meaningless reject x = 9 as an answer

This show that also the equation has a root that does not check because this equation does not  meet the hidden requirements that the rug have a positive length.  

Example 2
The FORMULA h = rt – 4.9t² is a good approximation of the height (h) in meters of an object t seconds after it is projected upward with an initial speed of r meters per second.

An arrow is shot upward with an initial speed of 34.3 m/s. When will it be at a height of 49 m?

let t = the  number of seconds  after being shot that the arrow is 49 m high.

Let h = the height of the arrow= 49 m

Let r = the initial speed = 34.3 m/s

Substitute in the formula

h = rt – 4.9t²

49 = 34.3t – 4.9t²

4.9t² – 34.3t + 49 = 0  THINK—GCF???

4.9(t² – 7t + 10 ) = 0

4.9(t -2)(t -5) = 0

Using the ZERO PRODUCTS PROPERTY

t = 2 and t = 5

Therefore the arrow is 49 m high both 2 seconds and 5 seconds after being shot… on its way up and on its way down!

Example 3

If a number is added to its square, the results is 56. Find the number

 Let x = the number

x² + x = 56

(x +8)(x - 7) = 0

 x = -8 and x = 7

Find two consecutive positive odd integers whose product is 143

Let x = the first positive odd integer

Let x + 2 = the 2^nd positive odd integer

x(x + 2) = 143

x² + 2x = 143

x² + 2x – 143 = 0

(x + 13)(x – 11) = 0

x = -13 and x = 11

But ask only for the positive integers so reject -13

The two integers are 11 and  13

Example 4

The sum of the squares of two consecutive negative odd integers is 290. Find the integers

let x = the 1^st negative odd integer

let x + 2 = the 2^nd negative odd integer

x² + (x + 2)² = 290

x² + x² + 4x + 4 = 290

2x² + 4x – 286 = 0  THINK  GCF!!!

2(x²+ 2x -143) = 0

2(x + 13)(x -11) = 0

x = -13 and x = 11

But it ask for the negative integers  so reject x = 11

The two negative integers are -13 and -11