aim: solving verbal problems using quadratic equations course: adv. alg. & trig aim: how to...
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Aim: Solving Verbal Problems using Quadratic Equations
Course: Adv. Alg. & Trig
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.
Aim: Solving Verbal Problems using Quadratic Equations
Course: Adv. Alg. & Trig
Quadratic Verbal Problems
The square of a number decreased by 4 times the number equals 21. Find the number.
Let n equal the numberThe square of the number - n2
Four times a number - 4nn2 - 4n = 21
1. Put in standard form
n2 - 4n - 21 = 02. Factor
(n - 7)(n + 3) = 03. Set each factor equal to zero and solve.
n + 3 = 0 n = -3n - 7 = 0 n = 7
Aim: Solving Verbal Problems using Quadratic Equations
Course: Adv. Alg. & Trig
Quadratic Verbal Problems
The product of two consecutive, positive, even integers is 80. Find the integers.
n - first of the consecutive, positive, even integers
n + 2 the next consecutive, positive, even integer
n(n + 2)1. Put in standard form
n2 + 2n = 80 n2 + 2n - 80 = 02. Factor
(n + 10)(n - 8) = 03. Set each factor equal to zero and solve.
n - 8 = 0n = 8
n + 10 = 0n = -10
810
= 80
Aim: Solving Verbal Problems using Quadratic Equations
Course: Adv. Alg. & Trig
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.
x
x = altitude (height)
x + 7
x + 7 = base
Area of parallelogram A = bh x(x + 7) = 30x2 + 7x = 30x2 + 7x - 30 = 0
x + 10 = 0x = -10
x - 3 = 0x = 3
x + 7 = 10ht.
base
Area = 30 (x + 10)(x - 3) = 0
Aim: Solving Verbal Problems using Quadratic Equations
Course: Adv. Alg. & Trig
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
A of rectangle = = 32
s = side of originals + 2 = new length of first pair of sides
s – 2 = new length of second pair of sides
(s + 2)(s – 2)
s2 – 4 = 32
s2 = 36 s = ±6
= 6
= 32(6 + 2)(6 – 2)8 · 4 = 32