Section 6.8
Number Problems
The product of two consecutive odd integers is 63.
Find the integers.
Plan:
Let x = the first odd integer;
then x + 2 = the second consecutive odd
integer.
An equation that describes the situation is:
x(x + 2) = 63
Carry Out:
x(x + 2) = 63
x2 + 2x = 63
x2 + 2x – 63 = 0
(x – 7) (x + 9) = 0
x – 7 = 0 or x + 9 = 0
x = 7 or
x = –9
If the first odd integer is 7, the next consecutive odd integer is
7 + 2 = 9.
If the first odd integer is –9, the next consecutive odd integer is
–9 + 2 = –7.
So there are two solutions (or two pairs of
consecutive odd integers whose product is 63).
Check:
There are two pairs of consecutive odd integers that
are solutions. They are 7, 9 and –9, –7.
 7, 9 is a solution because 7(9) = 63
 –9, –7 is a solution because –7(–9) = 63
ANSWER:
7, 9 and –9, –7
Geometry Problems
The length of a rectangle is 3 more than twice the
width.
The area is 44 square inches.
Find the dimensions (the length and width).
Plan:
Let x = the width of the rectangle.
Then 2x + 3 = the length of
the rectangle
Since the area is 44 square inches, an equation that
describes the situation is
Formula for Area of a
x(2x + 3) = 44
Rectangle
Carry Out:
x(2x + 3) = 44
2x2 + 3x = 44
2x2 + 3x – 44 = 0
(2x + 11)(x – 4)= 0
2x + 11 = 0
or x – 4 = 0
x=
or
x=4
A = lw
Length  width = area
The solution x =
cannot be used since length and
width are always given in positive units.
The width is 4.
The length is 3 more than twice the width or
2(4) + 3 = 11.
Check:
The solutions check in the original problem since
4(11) = 44.
The Answer:
Width = 4 inches
Length = 11 inches
The Pythagorean Theorem
The three sides of a right triangle are three
consecutive integers. Find the lengths of the three
sides.
Plan:
Let x = the first integer (shortest side)
then x + 1 = the next consecutive
integer
and x + 2 = the third consecutive
integer (longest side)
The Pythagorean theorem tells us that the square
of the longest side (x + 2)2 is equal to the sum of
the squares of the two shorter sides, (x + 1)2 + x2.
Here is the equation:
c2
=
a2
+ b2
(x + 2)2 = (x + 1)2 + x2
Carry Out:
(x + 2)2 = (x + 1)2 + x2
x2 + 4x + 4 = x2 + 2x + 1 + x2
Expand squares
x2 – 2x – 3 = 0
Standard form
(x – 3)(x + 1) = 0
x – 3 = 0 or
x = 3 or
x+1=0
x = –1
Factor
Set factors to 0
Since a triangle cannot have a side with a negative
number for its length, we can’t not use –1 for a
solution to our original problem; therefore, the
shortest side is 3.
The other two sides are the next two consecutive
integers, 4 and 5.
Check:
We can use the Pythagorean Theorem to check.
a b c
2
2
2
3 4 5
2
2
2
9 + 16 = 25
25 = 25



Section 6.8
Pages 475-479
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