Algebra (Period 1)

Using Equations That Factor 6-9 Cont'd

Word Problems continued...
We looked at a number of word problems... translated them to an equation and solved. The following should be in your spiral notebook:

1. The product of two consecutive positive even integers is 48. Find the two integers.

Let x = the first positive even integer
let x + 1 represent the second positive even integer

x(x +1) =48
x² + x = 48
move the 48 to the left side, setting the equation to zero

x² + x - 48 = 0
NOW factor
(x + 8)(x - 6) = 0
x = -8 x = 6
Stop and re-read the equation... It says positive even integers so the only solution is
x = 6
BUT... you need to find the two integers
x = 6
and x + 2 = 8
{6,8}

2. One side of a rectangle is 4 inches longer than the other. If the sides are each increased by 2 inches, the area of the new rectangle is 60 inches squared. How long are the sides of the original rectangle?

Let w = the width of the rectangle.
let w + 4 = the length of the rectangle. ( re-read the equation to get this information!!)
Then it says "sides are each increased by 2 inches"
so we have w + 2
and (w+4) + 2, which is just w + 6
Now the area is 60 inches squared so we know A = lw
(w +2)(w + 6) = 60
FOIL or use the box method...
w² + 8w + 12 = 60
We need to use the ZERO Products Property so we need to move the 60 to the left, setting the equation equal to ZERO
w² + 8w + 12 - 60 = 0
w² + 8w - 48 = 0
FACTOR
(w + 12)(w -4) = 0
w = -12 and w = 4
BUT A rectangle can't have a negative length so -12 is NOT a solution.

If w = 4
w + 4 = 8
The original rectangle's sides are 4 inches and 8 inches.

3. The sum of the squares of two consecutive even integers is 244. Find the two integers.
let n = the first even integer
let n + 2 = the second even integer

The sum of the squares---> n² + (n +2)² = 244
n² + n² + 4n + 4 = 244
combine like terms and move 244 to the left side of the equation you get
2n² + 4n - 240 = 0
Factor out the GCF
2(n² + 2n - 120) = 0
factor
2(n + 12)(n - 10) = 0
n = -12 and n = 10
This time both of those fit the requirements. Both are solutions BUT you need two sets of pairs.
n + 2 when n = -12 is - 12 + 2 = -10
and
n + 2 when n = 10 is 10 + 2 = 12

{-12, -10} and (10, 12)

4. The sum of twice a number and the number squared is -1. What is this number?
let x = a number
2x + x² = -1
2x + x² + 1 = 0
BUT WAIT... you can't work with this trinomial in this order... put in descending order and THEN factor..
x² + 2x + 1 = 0
Wait... do you see?... It's a Trinomial squared!!
(x + 1)² = 0
x = -1

5. Five times a number decreased by 6 is equal to the number squared. What is the number?

Let x = the number
5x - 6 = x ²
Keep the x² positive... that is move the other terms to it's side... easier to factor that way..
you get
0 = x² - 5x + 6
or
x² - 5x + 6 = 0
factor...
(x -3)(x -2) = 0
x = 3 and x = 2

We did a few more from our textbook on Page 294