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Quadratic Applications Help

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By — McGraw-Hill Professional
Updated on Sep 27, 2011

Introduction to Quadratic Application Word Problems

Most of the problems in this chapter are not much different from the word problems in previous chapters. The only difference is that quadratic equations are used to solve them. Because quadratic equations usually have two solutions, some of these applied problems will have two solutions. Most will have only one—one of the “solutions” will be invalid. More often than not, the invalid solutions are easy to recognize.

Solving for Consecutive Numbers

Examples

Example 1:

The product of two consecutive positive numbers is 240. Find the numbers.

Let x represent the first number. Because the numbers are consecutive, the next number is one more than the first: x + 1 represents the next number. The product of these two numbers is x ( x + 1), which equals 240.

 

x ( x + 1)

= 240

x 2 + x

= 240

x 2 + x – 240

= 0

( x – 15)( x + 16)

= 0

 

x – 15 = 0   ( x + 16 = 0 leads to a negative solution)

  +15  +15

x = 15

The consecutive positive numbers are 15 and 16.

(This problem could have been set up with x representing the first number and x – 1 representing the second number.)

Example 2:

The product of two consecutive even numbers is 528. What are the numbers?

Let x represent the first number. Consecutive even numbers (and consecutive odd numbers) differ by two, so let x + 2 represent the second number. Their product is x ( x + 2).

x ( x + 2)

= 528

x 2 + 2 x

= 528

x 2 + 2 x – 528

= 0

( x – 22)( x + 24)

= 0

 

Rational Equations Examples

The two solutions are 22 and 24, and –24 and –22.

Solving for Non-Consecutive Numbers Example

Example

Two positive numbers differ by five. Their product is 104. Find the two numbers.

Let x represent the first number. If x differs from the other number by five, then the other number could either be x + 5 or x – 5; it does not matter which representation you use. We will work this problem with both representations.

Let x + 5 represent the other number

Let x − 5 represent the other number

x ( x + 5) = 104

x ( x − 5) = 104

x 2 + 5 x = 104

x 2 − 5 x = 104

x 2 + 5 x − 104 = 0 ( x + 13)( x − 8) = 0 x − 8 = 0 ( x + 13 = 0 leads to a +8 +8 negative solution) x = 8 The numbers are 8 and 8 + 5 = 13.

x 2 - 5 x − 104 = 0 ( x - 13)( x + 8) = 0 x − 13 = 0 ( x + 8 = 0 leads to a +13 +13 negative solution) x = 13 The numbers are 13 and 13 − 5 = 8.

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