1. Aim: How to solve verbal problems with quadratic equations?
Do Now:
The square of a number decreased by 4 times the number equals 21. Find the number.

2. Quadratic Verbal Problems
The square of a number decreased by 4 times the number equals 21. Find the number.
Let n equal the number
The square of the number - n2
Four times a number - 4n
n2 - 4n = 21
1. Put in standard form
n2 - 4n - 21 = 0
2. Factor
(n - 7)(n + 3) = 0
3. Set each factor equal to zero and solve.
n + 3 = 0
n = -3
n - 7 = 0
n = 7

3. Quadratic Verbal Problems
The product of two consecutive, positive, even integers is 80. Find the integers. 8
n - first of the consecutive, positive, even integers 10
n + 2 the next consecutive, positive, even integer
n(n + 2)
= 80
1. Put in standard form
n2 + 2n = 80
n2 + 2n - 80 = 0
2. Factor
(n + 10)(n - 8) = 0
3. Set each factor equal to zero and solve.
n - 8 = 0
n = 8
n + 10 = 0
n = -10

4. Quadratic Verbal Problems
The base of a parallelogram measures 7 centimeters more than its altitude. If the area of the parallelogram is 30 square centimeters, find the measure of its base and the measure of its altitude.
Area of parallelogram A = bh
x(x + 7) = 30 x
x = altitude (height)
x2 + 7x = 30
x + 7
x + 7 = base
x2 + 7x - 30 = 0
Area = 30
(x + 10)(x - 3) = 0
x + 10 = 0
x = -10
x - 3 = 0
x = 3
ht.
x + 7 = 10
base

5. Quadratic Verbal Problems
The length of each of a pair of parallel sides of a square is increased by 2 meters, and the length of each of the other two sides of the square is decreased by 2 meters. The area of the rectangle formed is 32 square meters. Find the measure of one side of the original square.
Area of square A = s2
s = side of original
= 6
s + 2 = new length of first pair of sides
s – 2 = new length of second pair of sides
A of rectangle =
(s + 2)
(s – 2)
= 32
s2 – 4 = 32
= 32
(6 + 2)
(6 – 2)
8 · 4 = 32
s2 = 36
s = ±6