Solving $(x+3)^2 + (x+3) - 2 = 0$ A Step-by-Step Guide

Introduction

In this article, we will delve into the process of solving the quadratic equation $(x+3)^2 + (x+3) - 2 = 0$. This equation might look intimidating at first glance, but we can simplify it using a clever substitution technique. By recognizing the repeating term $(x+3)$, we can introduce a new variable, transforming the equation into a more manageable quadratic form. This approach not only makes the equation easier to solve but also provides a valuable strategy for tackling similar problems in algebra. Understanding how to manipulate and solve quadratic equations is fundamental in mathematics, with applications spanning various fields, including physics, engineering, and computer science. Therefore, mastering these techniques is crucial for anyone pursuing studies or careers in these areas. In the following sections, we will break down the solution step by step, ensuring a clear and comprehensive understanding of the method.

Substitution Method

The key to efficiently solving this equation lies in recognizing the repeated expression $(x+3)$. We can introduce a substitution to simplify the equation. Let's set $u = (x+3)$. This substitution transforms the original equation into a simpler quadratic equation in terms of $u$. By replacing $(x+3)$ with $u$, the equation $(x+3)^2 + (x+3) - 2 = 0$ becomes $u^2 + u - 2 = 0$. This new equation is a standard quadratic equation that we can solve using various methods, such as factoring, completing the square, or using the quadratic formula. Factoring is often the quickest method when the quadratic expression can be easily factored. In this case, we look for two numbers that multiply to -2 and add to 1. These numbers are 2 and -1. Thus, the quadratic expression can be factored as $(u+2)(u-1)=0$. This factored form allows us to easily find the values of $u$ that satisfy the equation. Once we have solved for $u$, we can substitute back the original expression $(x+3)$ to find the values of $x$. This substitution method is a powerful technique for simplifying equations and making them easier to solve. It is particularly useful when dealing with complex expressions or repeated terms within an equation. By mastering this method, you can significantly improve your problem-solving skills in algebra and beyond.

Solving for u

Now, let's proceed with solving the simplified quadratic equation $u^2 + u - 2 = 0$. As discussed in the previous section, we can factor this equation as $(u+2)(u-1) = 0$. This factorization is based on the principle that if the product of two factors is zero, then at least one of the factors must be zero. Therefore, we have two possible cases: $u+2=0$ or $u-1=0$. Solving the first case, $u+2=0$, we subtract 2 from both sides to get $u = -2$. Solving the second case, $u-1=0$, we add 1 to both sides to get $u = 1$. Thus, we have found two solutions for $u$: $u = -2$ and $u = 1$. These values of $u$ are the roots of the simplified quadratic equation. However, our original goal was to find the values of $x$ that satisfy the original equation. Therefore, we need to substitute back the expression for $u$ in terms of $x$ to find the corresponding values of $x$. This process of substituting back is crucial to completing the solution and answering the original question. In the next section, we will perform this substitution and solve for $x$.

Substituting Back and Solving for x

Having found the values of $u$, we now need to substitute back $u = (x+3)$ to solve for $x$. We have two cases to consider: $u = -2$ and $u = 1$. For the first case, where $u = -2$, we substitute this value into the expression $u = (x+3)$ to get $-2 = (x+3)$. To solve for $x$, we subtract 3 from both sides of the equation, resulting in $x = -2 - 3$, which simplifies to $x = -5$. For the second case, where $u = 1$, we substitute this value into the expression $u = (x+3)$ to get $1 = (x+3)$. To solve for $x$, we subtract 3 from both sides of the equation, resulting in $x = 1 - 3$, which simplifies to $x = -2$. Therefore, we have found two solutions for $x$: $x = -5$ and $x = -2$. These values of $x$ are the roots of the original equation $(x+3)^2 + (x+3) - 2 = 0$. To verify that these solutions are correct, we can substitute them back into the original equation and check if the equation holds true. This verification step is important to ensure the accuracy of our solutions. In the next section, we will perform this verification.

Verification

To ensure the correctness of our solutions, we will substitute the values $x = -5$ and $x = -2$ back into the original equation $(x+3)^2 + (x+3) - 2 = 0$. First, let's substitute $x = -5$ into the equation. We get $(-5+3)^2 + (-5+3) - 2 = (-2)^2 + (-2) - 2 = 4 - 2 - 2 = 0$. Since the equation holds true, $x = -5$ is indeed a solution. Next, let's substitute $x = -2$ into the equation. We get $(-2+3)^2 + (-2+3) - 2 = (1)^2 + (1) - 2 = 1 + 1 - 2 = 0$. Since the equation also holds true for $x = -2$, we have verified that both $x = -5$ and $x = -2$ are solutions to the original equation. This verification step is a crucial part of the problem-solving process, as it helps to catch any errors that may have occurred during the solution process. By verifying our solutions, we can be confident that our answers are correct. In conclusion, the solutions to the equation $(x+3)^2 + (x+3) - 2 = 0$ are $x = -5$ and $x = -2$.

Conclusion

In this article, we successfully solved the quadratic equation $(x+3)^2 + (x+3) - 2 = 0$ using the substitution method. We began by recognizing the repeating term $(x+3)$ and introducing a new variable $u = (x+3)$. This substitution transformed the original equation into a simpler quadratic equation, $u^2 + u - 2 = 0$. We then factored this equation as $(u+2)(u-1) = 0$, which led us to the solutions $u = -2$ and $u = 1$. Next, we substituted back the original expression for $u$ in terms of $x$ to find the values of $x$. This gave us two equations: $x+3 = -2$ and $x+3 = 1$, which we solved to find $x = -5$ and $x = -2$, respectively. Finally, we verified these solutions by substituting them back into the original equation, confirming that both values satisfy the equation. The substitution method is a powerful technique for simplifying equations and making them easier to solve. It is particularly useful when dealing with complex expressions or repeated terms within an equation. By mastering this method, you can significantly improve your problem-solving skills in algebra and beyond. Understanding how to solve quadratic equations is a fundamental skill in mathematics, with applications in various fields such as physics, engineering, and computer science. Therefore, practicing these techniques is essential for anyone pursuing studies or careers in these areas.