x + 2 - Algebra House

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Algebra 2
Section 3.7
Number Problems
1. Find three consecutive even integers so that the first integer
times the second integer is 24.
x = 1st
x + 2 = 2nd
x + 4 = 3rd
x(x + 2) = 24
x2 + 2x = 24
distributive property
-24 -24
x2 + 2x – 24 = 0
(x + 6)(x – 4) = 0
x+6=0
-6
or
-6
x = -6
x + 2 = -4
x + 4 = -2
standard form
factor into 2 binomials
x–4=0
set each factor equal to 0
+4 +4
or
x=4
x+2=6
x+4=8
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2. Find three consecutive integers so that the square of the first,
increased by the square of the third, is 100.
x = 1st
x + 1 = 2nd
x + 2 = 3rd
x2 + (x + 2)2 = 100
x2 + x2 + 4x + 4 = 100
2x2 + 4x + 4 = 100
(x + 2)2 = (x + 2)(x + 2) FOIL method
combine like terms
-100 -100
2x2 + 4x – 96 = 0
2(x2 + 2x – 48) = 0 factor 2 out
2(x + 8)(x – 6) = 0 factor into 2 binomials
x + 8 = 0 or x – 6 = 0 set each factor equal to 0
-8 -8
x = -8
x + 1 = -7
x + 2 = -6
+6 +6
or
x=6
x+1=7
x+2=8
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3. Forty chairs are placed in rows so that the number of chairs in
each row is 3 less than the number of rows. Find the number of
chairs in each row.
x = number of rows
x – 3 = number of chairs in each row
x(x – 3) = 40
x2 – 3x = 40
-40
# of rows × # of chairs in each row = total chairs
-40
x2 – 3x – 40 = 0 standard form
(x – 8)(x + 5) = 0 factor into 2 binomials
x – 8 = 0 or x + 5 = 0
+8 +8
-5 -5
x = -5
x = 8 rows
x – 3 = 5 chairs in each row
can’t have a negative number of rows
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4. One-hundred forty peaches were packed in some boxes so that the
number of boxes was 6 less than twice the number of peaches in
each box. Find the number of boxes used.
x = number of peaches in each box
2x – 6 = number of boxes
x(2x – 6) = 140
2x2 – 6x = 140
-140
# of peaches in each box × # of boxes = total peaches
-140
2x2 – 6x – 140 = 0
2(x2 – 3x – 70) = 0 factor 2 out
2(x – 10)(x + 7) = 0 factor into 2 binomials
x – 10 = 0 or x + 7 = 0
+10 +10
-7 -7
x = 10
x = -7
2x – 6 = 14 boxes used
can’t have a negative number of peaches
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Solve each problem.
Algebra 2
Section 3.7
Assignment
1.) Find three consecutive integers so that the product of the second and the
third integers is 42.
2.) Find three consecutive odd integers so that the sum of the squares of the
first two integers is 130.
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