## How to Solve Quadratic Word Problems Part 1

In the previous posts, we have learned how to solve quadratic equations by getting the extracting the square root, by factoring, and by quadratic formula. We continue this series by learning how to solve math word problems using quadratic equations. Most of the time, we need to rewrite the equation to the general form which is $ax^2 + bx + c = 0$.

Problem 1

The product of two consecutive positive even numbers is 48. What are the numbers?

Solution and Explanation

This problem can be solved mentally and by simple guess and check; however, we will solve it algebraically in order to illustrate the method of using quadratic equations.

Let $x$ = smaller number $x + 2$ = larger number

From the given above, we can form the following equation.

Smaller number times larger number = 48 $x(x + 2) = 48$

By the distributive property, this results to $x^2 + 2x = 48$

Now, we need to make this equation in general form $ax^2 + bx + c = 0$ so we can factor easily. To do this, we subtract 48 from both sides resulting to $x^2 + 2x - 48 = 0$

By factoring, we need two numbers whose sum is 2 and product is -48 where the absolute value of the larger number is greater than that of the smaller. With this restriction in mind, we have the following pairs of factors whose product is -48.

{48, -1}, {24, -2}, {12, -4}, {8,-6}

From these pairs, 8 and -6 has a sum of 2. Therefore, the factors are $(x + 8)(x - 6) = 0$

Equating to 0, we have $x + 8 = 0$, $x = -8$ $x - 6 = 0$, $x = 6$

Since we are looking for positive integers, we will take $x = 6$ and $x + 2 = 6 + 2 = 8$.

Therefore, the two consecutive numbers are 6 and 8.

Check: The numbers 6 and 8 are consecutive numbers and their product is 48. Therefore, we are correct.

## Introduction to Quadratic Equation

The length of a rectangle is 3 cm more than its width. Its area is equal to 54 square centimeters. What is its length and width?

Solution

Let

x = width of rectangle
x + 3 = length of rectangle

The area of a rectangle is the product of the length and width, so we have

Area= x(x + 3)

which is equal to 54.

Therefore, we can form the following equation:

x(x + 3) = 54.

By the distributive property, we have $x^2 + 3x = 54$

Finding the value of x

In the equation, we want to find the value of x that makes the equation true. Without algebraic manipulation, we can find the value of x by assigning various values to x. The equation $x^2 + 3x = 54$ indicates that one number is greater than the other by 3 and their product is 54. Examining the numbers with product as 54, we have,

1 and 54
2 and 27
3 and 18
6 and 9.

Note: We have excluded the negative (e.g. (-1)(-54) = 54) numbers since a side length cannot be negative.

Now, 9-6 = 3 which means that the side lengths of the rectangle are 6 and 9. Yes, their product is 54 and one is 3 greater than the other.

In the equation above, subtracting both sides by 54, we have $x^2 + 3x - 54 = 54 - 54$ $x^2 + 3x - 54 = 0$.

The equation that we formed above is an example of a quadratic equation.

A quadratic equation is of the form $ax^2 + bx + c = 0$, where a, b, and c are real numbers and a not equal to 0. In the example above, a = 1, b = 3, and c = -54.

In the problem above, we got the value of x by testing several values, however, there are more systematic methods. In the next post, we will be discussing one of these methods. These methods are factoring, completing the square, and quadratic formula.

## How to Solve Rectangle Area Problems Part 2

We have already learned the concept of area of a rectangle and solved sample problems about it. In this post, we continue the rectangle area problems series. We discuss three more problems about rectangle area.

The fourth problem below involves area preservation, the fifth is calculating the area given its perimeter, and the sixth requiring the use of quadratic equations.

Let’s begin.

Problem 4

What is the area of the figure below?  Continue Reading