Aim: Solving Verbal Problems using Quadratic Equations

Course: Adv. Alg. & Trig

Aim: How to solve verbal problems with quadratic equations?

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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