Solving Quadratic Equations: A Case Study on a Special Number

Algebra is a powerful tool in mathematics that involves solving various types of equations, including quadratic equations. In this article, we will explore a specific problem that involves finding a number based on a quadratic relationship. Specifically, we will solve the problem, The square of a certain number is 22 less than 13 times the original number. This article will guide you through the process step-by-step, providing clarity and insight into the solution.

Problem Statement

The problem statement is as follows: The square of a certain number is 22 less than 13 times the original number. What is the number?

Formulating the Equation

Let the unknown number be x. According to the problem, we can set up the equation:

x2 13x - 22

Rearranging and Factoring

To solve for x, we start by rearranging the equation:

x2 - 13x - 22 0

This is a quadratic equation in the standard form ax2 bx c 0, where a 1, b -13, and c -22.

We can factor the quadratic equation. We need to find two numbers that multiply to -22 and add up to -13. The factors that fit are -11 and -2. Therefore, we can rewrite the equation as:

(x - 11)(x - 2) 0

Solving for x

Setting each factor equal to zero gives us the possible solutions:

x - 11 0 rArr; x 11

x - 2 0 rArr; x 2

The two possible numbers are 11 and 2.

Verification

To ensure our solutions are correct, we can check both values:

For x 11: x2 112 121 13x - 22 13(11) - 22 143 - 22 121 This is correct. For x 2: x2 22 4 13x - 22 13(2) - 22 26 - 22 4 This is also correct.

Thus, the numbers that satisfy the condition are boxed{11} and boxed{2}.

Conclusion

In conclusion, the quadratic equation was solved both algebraically and verified through substitution. The two possible numbers, 2 and 11, are the solutions to the problem. Such exercises are crucial in developing skills in algebra and understanding the application of quadratic equations in solving real-world problems.