Solve Quadratic Equation: $(5x+5)^2+8(5x+5)+16=0$

When faced with a mathematical equation, especially one that looks a bit complex like $(5x+5)^2+8(5x+5)+16=0$, it's natural to feel a slight pause. However, with a systematic approach, we can unravel its complexities and find the real solutions. This equation might seem intimidating at first glance due to the repeated term $(5x+5)$, but this repetition is actually a key to simplifying it. We'll dive deep into techniques that make solving such equations straightforward and even enjoyable for anyone interested in the beauty of mathematics. Understanding how to tackle these problems is a fundamental skill that opens doors to more advanced algebraic concepts and real-world applications. So, let's embark on this mathematical journey together and discover the elegance of finding solutions.

Understanding the Structure of the Equation

Let's first take a good look at the equation: $(5x+5)^2+8(5x+5)+16=0$. The immediate thing that catches the eye is the recurring expression $(5x+5)$. This isn't just a coincidence; it's a deliberate structure that hints at a powerful simplification technique. Instead of expanding $(5x+5)^2$ and then distributing the 8, which would lead to a much larger and more complicated quadratic equation, we can use a substitution. This method is often referred to as a 'strategic substitution' or 'change of variable'. By recognizing the repeating part, we can rename it to something simpler, say, 'u'. So, let $u = 5x+5$. Once we make this substitution, the original equation transforms into a much more familiar form: $u^2 + 8u + 16 = 0$. This new equation is a standard quadratic equation in terms of 'u', and solving it is a well-established procedure in algebra. This transformation is a classic example of how recognizing patterns can dramatically simplify problem-solving in mathematics. It's like finding a shortcut on a long journey; the destination remains the same, but the path becomes much easier. The principle behind this substitution is that it reduces the complexity of the expression, making it easier to apply standard algebraic techniques. We are essentially abstracting away the more complex part of the equation, dealing with a simpler form, and then translating the solution back to the original variable. This approach is not only efficient but also demonstrates a deeper understanding of algebraic manipulation. It’s a testament to the power of abstraction in mathematics, where complex problems can often be broken down into simpler, manageable components.

Applying Substitution for Simplification

Now that we've established the substitution $u = 5x+5$, our equation becomes $u^2 + 8u + 16 = 0$. This is a quadratic equation in its most basic form. Our goal is to find the values of 'u' that satisfy this equation. There are several methods to solve quadratic equations, including factoring, using the quadratic formula, or completing the square. Let's consider factoring first, as it's often the quickest method if the expression is easily factorable. We are looking for two numbers that multiply to 16 (the constant term) and add up to 8 (the coefficient of the 'u' term). The numbers 4 and 4 fit this description perfectly: $4 imes 4 = 16$ and $4 + 4 = 8$. Therefore, we can factor the quadratic equation as $(u+4)(u+4) = 0$, which simplifies to $(u+4)^2 = 0$. This is a perfect square trinomial. For the product of two terms to be zero, at least one of the terms must be zero. In this case, since both terms are identical, we have $u+4 = 0$. Solving for 'u', we get $u = -4$. This is the unique solution for 'u'. The fact that we have a squared term equaling zero indicates that there is exactly one real solution for 'u', a concept known as a repeated root or a double root. This simplification through substitution is a cornerstone of algebraic problem-solving, allowing us to transform intricate expressions into manageable forms. It highlights the elegance of mathematical structures, where a seemingly complex problem can be reduced to a simple one by observing and utilizing its inherent patterns. The process of factoring a perfect square trinomial is a key skill, and recognizing it can save significant time and effort in solving equations. This particular equation is a textbook example of how substitution can elegantly reveal the underlying simplicity of a problem, making it accessible and solvable.

Substituting Back to Find 'x'

We have successfully found the value of 'u', which is $u = -4$. However, our original goal was to find the value(s) of 'x'. Remember that we made the substitution $u = 5x+5$. Now, we need to substitute the value of 'u' back into this expression to solve for 'x'. So, we set $-4 = 5x+5$. This is a simple linear equation, which is much easier to solve than the original quadratic equation. To isolate 'x', we first subtract 5 from both sides of the equation: $-4 - 5 = 5x$. This gives us $-9 = 5x$. Now, to find 'x', we divide both sides by 5: $x = -9/5$. Thus, the real solution to the original equation $(5x+5)^2+8(5x+5)+16=0$ is $x = -9/5$. It's important to note that because the quadratic equation in 'u' had a repeated root (a single solution for 'u'), the original equation in 'x' also has a single real solution. If the quadratic in 'u' had yielded two different values for 'u', we would have had two different values for 'x' to find. The process of substituting back is crucial; it connects the simplified problem back to the original context, providing the answer in terms of the initial variable. This step underscores the importance of keeping track of substitutions made and their corresponding original expressions. It's the final bridge that leads us from the abstract solution of 'u' to the concrete solution for 'x', completing the problem-solving journey. This methodical approach ensures that we don't miss any steps and that our final answer is indeed the solution to the equation we started with, demonstrating the power of reversing the substitution process.

Verifying the Solution

To ensure that our solution $x = -9/5$ is correct, we can substitute it back into the original equation $(5x+5)^2+8(5x+5)+16=0$. Let's first calculate the value of $5x+5$ when $x = -9/5$: $5(-9/5) + 5 = -9 + 5 = -4$. Now, substitute this value back into the equation: $(-4)^2 + 8(-4) + 16 = 0$. Let's evaluate this: $16 - 32 + 16 = 0$. Simplifying further, $32 - 32 = 0$. Since $0 = 0$, our solution is verified. This verification step is a critical part of the problem-solving process in mathematics. It not only confirms the accuracy of our answer but also reinforces our understanding of how the equation works. By plugging the obtained value of 'x' back into the original equation and seeing that it holds true, we gain confidence in our algebraic manipulations and the methods used. This process is like checking your work after completing a complex task; it helps catch any potential errors and ensures the integrity of the final result. In more complex problems, especially those involving multiple steps or advanced concepts, verification can be invaluable in identifying mistakes that might have been made along the way. It’s a practice that builds rigor and precision into mathematical work, making the solutions more reliable and trustworthy. This systematic confirmation is a hallmark of good mathematical practice, ensuring that the solutions we arrive at are not just plausible but demonstrably correct according to the rules of algebra. It's a satisfying final step that brings closure to the problem.

Alternative Method: Direct Expansion (and why it's less ideal)

While substitution is the most efficient way to solve this particular equation, it's worth briefly exploring what would happen if we chose to expand everything directly. This approach, while longer, can sometimes be a useful way to understand why substitution is preferred. Expanding $(5x+5)^2$ gives us $(5x)^2 + 2(5x)(5) + 5^2 = 25x^2 + 50x + 25$. Distributing the 8 to $(5x+5)$ gives us $8(5x+5) = 40x + 40$. Now, substituting these back into the original equation, we get: $(25x^2 + 50x + 25) + (40x + 40) + 16 = 0$. Combining like terms, we have $25x^2 + (50x + 40x) + (25 + 40 + 16) = 0$. This simplifies to $25x^2 + 90x + 81 = 0$. Now we have a standard quadratic equation in the form $ax^2 + bx + c = 0$, where $a=25$, $b=90$, and $c=81$. We could solve this using the quadratic formula x = rac{-b andpm ext{sqrt}(b^2 - 4ac)}{2a}. Let's calculate the discriminant, anddelta = b^2 - 4ac: anddelta = (90)^2 - 4(25)(81) = 8100 - 100(81) = 8100 - 8100 = 0. Since the discriminant is 0, there is exactly one real solution. Using the quadratic formula: x = rac{-90 andpm ext{sqrt}(0)}{2(25)} = rac{-90}{50} = -rac{9}{5}. As you can see, we arrive at the same solution, $x = -9/5$. However, the direct expansion involved much more arithmetic and a higher chance of making calculation errors compared to the substitution method. This illustrates the power and elegance of using substitutions to simplify complex algebraic expressions, turning a potentially tedious calculation into a more straightforward process. It highlights that in mathematics, recognizing patterns and applying appropriate techniques can significantly streamline problem-solving.

Conclusion: The Power of Strategic Simplification

In conclusion, solving the equation $(5x+5)^2+8(5x+5)+16=0$ beautifully demonstrates the power of strategic simplification in mathematics. By recognizing the repeating term $(5x+5)$ and employing a substitution ($u = 5x+5$), we transformed a complex-looking equation into a simple quadratic $u^2 + 8u + 16 = 0$. This new form was easily factorable into $(u+4)^2 = 0$, yielding a single solution for $u$, which is $u = -4$. The subsequent step of substituting back $5x+5$ for $u$ led us to the linear equation $-4 = 5x+5$, from which we derived the unique real solution for $x$ as $x = -9/5$. The verification step confirmed the accuracy of our solution. We also saw that direct expansion, while yielding the same result, was a more cumbersome and error-prone process. This problem serves as an excellent reminder that taking a moment to observe the structure of an equation can often reveal elegant shortcuts. Mastering techniques like substitution not only makes solving algebraic problems more efficient but also deepens our appreciation for the underlying logic and beauty of mathematics. It’s a foundational skill that prepares us for more challenging mathematical endeavors.

For further exploration into solving algebraic equations and understanding quadratic functions, you can visit Khan Academy or Math is Fun.