# 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.