Solve $(a+4)^2+3(a+4)+2=0$ A Step-by-Step Guide

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In this comprehensive guide, we will delve into the intricacies of solving the quadratic equation $(a+4)^2+3(a+4)+2=0$. This equation, while appearing complex at first glance, can be efficiently tackled using various algebraic techniques. Our primary goal is to not only find the solutions for a but also to thoroughly check these solutions to ensure their validity. This process of solving and checking is crucial in mathematics to guarantee the accuracy and correctness of our results. Throughout this article, we will emphasize clarity and precision, providing step-by-step explanations that will empower you to confidently solve similar problems. Whether you are a student seeking to enhance your algebra skills or a mathematics enthusiast eager to deepen your understanding, this guide will offer valuable insights and practical techniques. Let's embark on this mathematical journey together, transforming complexity into clarity and mastering the art of solving quadratic equations.

Understanding the Quadratic Equation

Before we jump into solving the specific equation $(a+4)^2+3(a+4)+2=0$, it's essential to grasp the fundamental concept of a quadratic equation. A quadratic equation is a polynomial equation of the second degree. The general form of a quadratic equation is given by:

$ax^2 + bx + c = 0$

where a, b, and c are constants, and x represents the variable. The coefficient a must not be equal to zero, as this would reduce the equation to a linear form. Understanding this general form is the cornerstone to tackling any quadratic equation, and it's the first step in demystifying the complexities that might initially seem daunting. When we examine our equation, $(a+4)^2+3(a+4)+2=0$, we can observe that it possesses a structure similar to the general quadratic form, albeit with a slight twist. The term $(a+4)$ acts as a composite variable, which we can simplify to make the equation more recognizable and easier to solve. This ability to recognize patterns and adapt standard forms is a critical skill in mathematics, allowing us to approach a wide range of problems with confidence and flexibility. So, as we delve deeper into the solution process, remember that our foundation lies in the understanding of the quadratic equation's general form and how it manifests in different scenarios. The power of algebra lies in its ability to transform seemingly complex expressions into manageable forms, and this equation is no exception. By breaking it down step by step, we'll uncover its underlying structure and reveal the path to its solutions.

Method 1: Substitution

One of the most effective strategies for solving equations like $(a+4)^2+3(a+4)+2=0$ is the method of substitution. This technique involves replacing a complex expression with a single variable to simplify the equation. In our case, we can substitute $u = (a+4)$. By doing this, we transform the original equation into a more familiar quadratic form, which is easier to solve. The substitution method is not just a shortcut; it's a powerful tool that allows us to break down intricate problems into manageable parts. It’s a testament to the elegance of mathematical problem-solving, where a clever substitution can often unveil the underlying simplicity of an equation. Now, with the substitution $u = (a+4)$, our equation becomes:

$u^2 + 3u + 2 = 0$

This equation is a standard quadratic equation, which we can solve using various methods, such as factoring, completing the square, or the quadratic formula. Factoring, in this case, is the most straightforward approach. We look for two numbers that multiply to 2 and add up to 3. These numbers are 1 and 2. Therefore, we can factor the quadratic expression as:

$(u + 1)(u + 2) = 0$

This factored form gives us two possible solutions for u: $u = -1$ and $u = -2$. However, we are not done yet. We need to find the values of a, not u. This is where the beauty of substitution comes full circle. We now substitute back $(a+4)$ for u in each solution, which will lead us to the values of a that satisfy the original equation. This back-substitution is a crucial step, reminding us that our ultimate goal is to solve for the original variable. The elegance of the substitution method lies not only in its ability to simplify complex expressions but also in its systematic approach, ensuring that we don't lose sight of the original problem. As we proceed, we'll see how this method unravels the complexities of the equation, revealing the solutions with clarity and precision.

Solving for u

Having substituted $u = (a+4)$ into our original equation, we arrived at the simplified quadratic form:

$u^2 + 3u + 2 = 0$

Now, the task at hand is to solve for u. As mentioned earlier, factoring is a particularly efficient method for this equation. Factoring involves expressing the quadratic expression as a product of two binomials. To factor $u^2 + 3u + 2$, we need to find two numbers that multiply to the constant term (2) and add up to the coefficient of the linear term (3). After a bit of mental calculation or a quick examination of factor pairs, we find that the numbers 1 and 2 satisfy these conditions, as 1 * 2 = 2 and 1 + 2 = 3. With these numbers in hand, we can rewrite the quadratic equation in its factored form:

$(u + 1)(u + 2) = 0$

This factored form is a significant milestone in our solution process. It transforms a quadratic equation into a product of two expressions, each of which can be easily analyzed. The fundamental principle at play here is the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. In other words, if AB = 0, then either A = 0 or B = 0 (or both). Applying this property to our factored equation, we set each factor equal to zero:

$u + 1 = 0$ or $u + 2 = 0$

Solving these simple linear equations yields two values for u:

$u = -1$ or $u = -2$

These are the solutions for u, but remember, our ultimate goal is to find the solutions for a. We have reached a crucial juncture where we must remember the substitution we made earlier. The values of u are stepping stones, guiding us back to the original variable. The next step involves substituting back $(a+4)$ for u in each of these solutions, a process that will reveal the corresponding values of a. This seamless transition between variables is a hallmark of algebraic problem-solving, highlighting the interconnectedness of different parts of a mathematical expression. As we continue, we will see how these values of u serve as a bridge, connecting us to the final solutions for a.

Substituting Back for a

Now that we have found the values of u to be -1 and -2, the next crucial step is to substitute back for a. Recall that we initially set $u = (a+4)$. This substitution simplified our equation and allowed us to solve for u, but our ultimate goal is to find the values of a that satisfy the original equation. To do this, we replace u with $(a+4)$ in each of our solutions for u. Let's start with the first solution, $u = -1$. Substituting $(a+4)$ for u, we get:

$a + 4 = -1$

This is a simple linear equation that we can solve for a by subtracting 4 from both sides:

$a = -1 - 4$

$a = -5$

So, one solution for a is -5. Now, let's move on to the second solution, $u = -2$. Substituting $(a+4)$ for u in this case gives us:

$a + 4 = -2$

Again, we solve for a by subtracting 4 from both sides:

$a = -2 - 4$

$a = -6$

Thus, the second solution for a is -6. We have now found two potential solutions for a: -5 and -6. However, in mathematics, finding solutions is only half the battle. It's equally important to check these solutions to ensure that they satisfy the original equation. This verification process is a cornerstone of mathematical rigor, safeguarding against errors and ensuring the validity of our results. As we proceed to check our solutions, we will reinforce our understanding of the equation and gain confidence in our solution process. Checking our answers is not just a formality; it's an integral part of the problem-solving journey, transforming potential solutions into confirmed answers.

Method 2: Expanding and Factoring

Another approach to solving the equation $(a+4)^2+3(a+4)+2=0$ involves expanding and factoring. This method provides a direct way to transform the equation into a standard quadratic form, which can then be factored to find the solutions. Expanding the equation means removing the parentheses and simplifying the expression. This process involves applying the distributive property and combining like terms. While it may seem more laborious than the substitution method, expanding and factoring is a fundamental algebraic technique that is widely applicable across various mathematical contexts. It's a testament to the versatility of algebra, where different approaches can lead to the same destination. The key to success with this method lies in careful execution and attention to detail. Each step, from expanding the squared term to combining constants, must be performed accurately to avoid errors. So, as we embark on this method, we'll emphasize precision and systematic thinking, ensuring that each step is a building block towards the final solution. This method not only offers an alternative route to the answer but also reinforces our understanding of algebraic manipulation and equation solving. The journey of expanding and factoring is a valuable exercise in mathematical discipline, honing our skills and deepening our appreciation for the structure of algebraic expressions. Let’s begin by expanding the equation and see how this process unfolds.

Expanding the Equation

To begin the process of solving $(a+4)^2+3(a+4)+2=0$ by expanding and factoring, we first need to expand the equation. This involves removing the parentheses and simplifying the expression. The first term we need to expand is $(a+4)^2$. This is a binomial squared, which can be expanded using the formula $(x+y)^2 = x^2 + 2xy + y^2$. Applying this formula to our term, we get:

$(a+4)^2 = a^2 + 2(a)(4) + 4^2 = a^2 + 8a + 16$

Now, let's expand the second term, $3(a+4)$. This involves distributing the 3 across the terms inside the parentheses:

$3(a+4) = 3a + 12$

With both terms expanded, we can now rewrite the original equation as:

$a^2 + 8a + 16 + 3a + 12 + 2 = 0$

The next step is to combine like terms. Like terms are terms that have the same variable raised to the same power. In this equation, we have the quadratic term $a^2$, linear terms $8a$ and $3a$, and constant terms 16, 12, and 2. Combining these terms, we get:

$a^2 + (8a + 3a) + (16 + 12 + 2) = 0$

Simplifying further, we have:

$a^2 + 11a + 30 = 0$

This resulting equation is a standard quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = 11$, and $c = 30$. This form is much easier to work with and allows us to apply various techniques for solving quadratic equations, such as factoring, completing the square, or the quadratic formula. We have successfully transformed the original equation into a more manageable form through expansion and simplification. The next step is to factor this quadratic equation, which will lead us to the solutions for a. The process of expanding the equation has not only simplified its form but also unveiled its underlying structure, setting the stage for the next phase of our solution journey. As we proceed to factor, we'll see how this transformation brings us closer to the values of a that satisfy the equation.

Factoring the Quadratic

Having expanded and simplified the original equation, we now have the quadratic equation:

$a^2 + 11a + 30 = 0$

The next step is to factor this quadratic. Factoring involves expressing the quadratic expression as a product of two binomials. To factor $a^2 + 11a + 30$, we need to find two numbers that multiply to the constant term (30) and add up to the coefficient of the linear term (11). This is a classic factoring puzzle, where the right combination of numbers unlocks the solution. We can systematically consider pairs of factors of 30: 1 and 30, 2 and 15, 3 and 10, 5 and 6. Among these pairs, 5 and 6 add up to 11, which is exactly what we need. Therefore, we can factor the quadratic expression as:

$(a + 5)(a + 6) = 0$

This factored form is a significant step forward. It transforms a single quadratic equation into a product of two binomial expressions, each of which can be analyzed independently. The underlying principle that allows us to proceed is the zero-product property, which we encountered earlier. This property states that if the product of two factors is zero, then at least one of the factors must be zero. In other words, if AB = 0, then either A = 0 or B = 0 (or both). Applying this property to our factored equation, we set each factor equal to zero:

$a + 5 = 0$ or $a + 6 = 0$

These are now simple linear equations that can be easily solved for a. Solving the first equation, $a + 5 = 0$, we subtract 5 from both sides to get:

$a = -5$

Similarly, solving the second equation, $a + 6 = 0$, we subtract 6 from both sides to get:

$a = -6$

Thus, we have found two potential solutions for a: -5 and -6. However, as we emphasized earlier, finding solutions is only part of the process. The next crucial step is to check these solutions in the original equation to ensure their validity. This verification step is not just a formality; it's an essential safeguard against errors and a testament to mathematical rigor. As we proceed to check our solutions, we'll reinforce our understanding of the equation and gain confidence in our results. Checking our answers transforms potential solutions into confirmed solutions, completing the puzzle and affirming the correctness of our method.

Checking the Solutions

After solving for a using two different methods, we have found two potential solutions: $a = -5$ and $a = -6$. However, in mathematics, it's crucial to check our solutions to ensure they are correct. This process involves substituting each solution back into the original equation and verifying that it satisfies the equation. Checking solutions is not just a formality; it's a critical step in the problem-solving process that helps us catch any errors we may have made along the way. It’s a testament to the rigor and precision that are hallmarks of mathematical thinking. By verifying our solutions, we not only confirm their correctness but also deepen our understanding of the equation itself. The act of substitution reinforces the relationship between the variables and the equation, solidifying our grasp of the underlying mathematical principles. Moreover, checking our solutions provides a sense of closure and confidence in our results. It transforms potential answers into confirmed solutions, completing the mathematical journey. So, let's embark on this crucial step, substituting each value of a back into the original equation and verifying that it holds true. This process will not only validate our solutions but also underscore the importance of accuracy and precision in mathematical problem-solving. Let’s begin by checking the first solution, $a = -5$, and then move on to the second solution, $a = -6$.

Checking $a = -5$

To check if $a = -5$ is a valid solution, we need to substitute this value back into the original equation:

$(a+4)^2+3(a+4)+2=0$

Replacing a with -5, we get:

$(-5+4)^2+3(-5+4)+2=0$

Now, let's simplify the expression step by step. First, simplify the terms inside the parentheses:

$(-1)^2 + 3(-1) + 2 = 0$

Next, evaluate the square and the multiplication:

$1 - 3 + 2 = 0$

Finally, perform the addition and subtraction:

$0 = 0$

Since the equation holds true when $a = -5$, we can confidently say that $a = -5$ is indeed a solution to the original equation. This verification step is a small but significant victory. It confirms that our algebraic manipulations and reasoning have been accurate. The fact that the equation balances when we substitute $a = -5$ provides assurance that we are on the right track. However, our journey is not yet complete. We have one more potential solution to check, $a = -6$. The process will be the same: substitute the value into the original equation and verify that it holds true. This systematic approach to checking solutions underscores the importance of thoroughness in mathematics. Each solution must be rigorously tested to ensure its validity. As we move on to check $a = -6$, we carry with us the confidence gained from verifying $a = -5$. This step-by-step approach, combined with careful attention to detail, is the key to mastering mathematical problem-solving. So, let's proceed to the next verification, knowing that each successful check brings us closer to a complete and accurate solution.

Checking $a = -6$

Now, let's check the second potential solution, $a = -6$. As we did with $a = -5$, we need to substitute this value back into the original equation:

$(a+4)^2+3(a+4)+2=0$

Replacing a with -6, we get:

$(-6+4)^2+3(-6+4)+2=0$

Now, let's simplify the expression step by step. First, simplify the terms inside the parentheses:

$(-2)^2 + 3(-2) + 2 = 0$

Next, evaluate the square and the multiplication:

$4 - 6 + 2 = 0$

Finally, perform the addition and subtraction:

$0 = 0$

Since the equation holds true when $a = -6$, we can confidently say that $a = -6$ is also a solution to the original equation. This second successful verification reinforces the accuracy of our solution process. It confirms that both values we found, $a = -5$ and $a = -6$, are indeed solutions to the quadratic equation. The fact that both solutions satisfy the equation underscores the completeness of our analysis. We have not only found the solutions but also rigorously verified them, leaving no room for doubt. This thorough approach is a hallmark of mathematical excellence, demonstrating a commitment to accuracy and precision. The process of checking solutions is more than just a formality; it’s a crucial step that transforms potential answers into confirmed results. It provides a sense of closure and confidence, knowing that we have solved the problem completely and correctly. With both solutions verified, we can now confidently state the final answer, knowing that it is supported by a rigorous and systematic solution process. So, as we conclude this problem, let's carry forward the lessons learned about careful algebraic manipulation, the importance of checking solutions, and the power of a step-by-step approach in mastering mathematical challenges. These skills will serve us well in tackling future problems and deepening our understanding of the mathematical world.

Final Answer

Having meticulously solved the equation $(a+4)^2+3(a+4)+2=0$ using two different methods and rigorously checked our solutions, we have arrived at the final answer. The solutions for a that satisfy the equation are:

$a = -5, -6$

These two values are the roots of the quadratic equation, representing the points where the equation equals zero. The journey to this final answer has been a testament to the power of algebraic techniques and the importance of careful, systematic problem-solving. We began by understanding the structure of the equation, recognizing it as a quadratic form with a composite variable. We then employed two distinct methods to unravel its complexities: substitution and expanding and factoring. The substitution method allowed us to simplify the equation by replacing $(a+4)$ with a single variable, transforming it into a more familiar quadratic form. This approach highlighted the elegance of algebraic manipulation, where a clever substitution can often reveal the underlying simplicity of a problem. The expanding and factoring method, on the other hand, provided a direct route to the solution by transforming the equation into a standard quadratic form through expansion and simplification. This method underscored the importance of precision and attention to detail, as each step required careful execution to avoid errors. Regardless of the method used, the final step of checking our solutions proved to be crucial. This verification process not only confirmed the correctness of our answers but also reinforced our understanding of the equation and the relationship between its variables. In conclusion, the solution to the equation $(a+4)^2+3(a+4)+2=0$ is $a = -5, -6$. This answer is not just a set of numbers; it's the culmination of a rigorous and systematic mathematical exploration, a journey that has honed our problem-solving skills and deepened our appreciation for the beauty and precision of algebra.