Solving (5x - 4)^2 + 10(5x - 4) + 25 = 0 A Step-by-Step Guide
This article aims to provide a comprehensive guide on how to find the real solutions of the given equation: (5x - 4)^2 + 10(5x - 4) + 25 = 0. We will walk through the process step-by-step, starting with rewriting the equation in quadratic form using an appropriate substitution. We'll cover the fundamental concepts of quadratic equations, methods for solving them, and ultimately arrive at the solutions for the given problem. Whether you're a student tackling algebra problems or someone looking to refresh your math skills, this guide will provide a clear and concise explanation. Let's dive in and unlock the secrets of this equation!
Understanding Quadratic Equations
Quadratic equations are polynomial equations of the second degree, meaning the highest power of the variable is 2. The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants, and a ≠ 0. These equations are fundamental in mathematics and have applications in various fields, including physics, engineering, and economics. Solving quadratic equations involves finding the values of the variable (usually 'x') that satisfy the equation. These values are also known as the roots or solutions of the equation. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. Each method has its advantages and disadvantages, and the choice of method often depends on the specific equation. In this article, we will focus on a technique that involves substitution to simplify the given equation into a more recognizable quadratic form.
Before we delve into the specifics of solving our equation, it's crucial to understand the core principles of quadratic equations. A quadratic equation is essentially a polynomial equation where the highest power of the unknown variable is 2. The most common form of a quadratic equation is expressed as ax^2 + bx + c = 0, where 'a', 'b', and 'c' represent constant coefficients, and 'x' is the variable we aim to solve for. The coefficient 'a' plays a vital role as it cannot be zero; otherwise, the equation would degenerate into a linear equation. Quadratic equations are not just abstract mathematical constructs; they have a wide range of applications in real-world scenarios. From determining the trajectory of a projectile in physics to modeling economic trends, quadratic equations provide powerful tools for understanding and predicting various phenomena. The solutions to a quadratic equation, often called roots or zeros, represent the values of 'x' that satisfy the equation. These solutions can be real or complex numbers, depending on the discriminant (b^2 - 4ac) of the equation. Understanding the nature of the roots is crucial in many applications. For instance, in engineering, real roots might represent physical dimensions, while complex roots might indicate instability in a system. The process of solving a quadratic equation involves finding these roots. There are several methods available, each with its own advantages and limitations. Factoring, completing the square, and using the quadratic formula are among the most common techniques. The choice of method often depends on the specific form of the equation and the ease with which it can be applied. In the context of the given equation, (5x - 4)^2 + 10(5x - 4) + 25 = 0, we will employ a technique known as substitution to transform it into a standard quadratic form. This approach simplifies the equation and allows us to apply familiar methods for finding the solutions. By understanding the fundamentals of quadratic equations, we can effectively tackle a wide range of problems and gain a deeper appreciation for their significance in mathematics and beyond.
Rewriting the Equation Using Substitution
The given equation, (5x - 4)^2 + 10(5x - 4) + 25 = 0, might seem complex at first glance. However, we can simplify it by using a substitution. The key is to recognize the repeating expression, which in this case is (5x - 4). By substituting this expression with a single variable, we can transform the equation into a more manageable quadratic form. Let's introduce a new variable, say 'u', and let u = (5x - 4). Now, we can rewrite the original equation in terms of 'u'. Substituting (5x - 4) with 'u', we get u^2 + 10u + 25 = 0. This new equation is a standard quadratic equation in the variable 'u'. It is in the form au^2 + bu + c = 0, where a = 1, b = 10, and c = 25. This form is much easier to work with than the original equation. By making this substitution, we have effectively transformed a seemingly complicated equation into a familiar quadratic form. This is a common technique used in algebra to simplify equations and make them easier to solve. The next step is to solve this quadratic equation for 'u'. Once we find the value(s) of 'u', we can substitute back to find the value(s) of 'x' that satisfy the original equation. The substitution method is a powerful tool for simplifying equations, and it is particularly useful when dealing with equations that contain repeating expressions. By making a suitable substitution, we can often transform a complex equation into a simpler one that we can solve using standard techniques.
The process of rewriting the equation using a suitable substitution is a crucial step in simplifying complex algebraic expressions. In the given equation, (5x - 4)^2 + 10(5x - 4) + 25 = 0, the repeating term (5x - 4) provides an excellent opportunity for substitution. By recognizing this pattern, we can introduce a new variable, say 'u', to represent (5x - 4). This substitution transforms the original equation into a more manageable quadratic form. Let's define u = (5x - 4). Now, we replace every instance of (5x - 4) in the original equation with 'u'. This gives us the new equation: u^2 + 10u + 25 = 0. This equation is a standard quadratic equation in terms of 'u', which is significantly easier to work with than the original equation in terms of 'x'. The coefficients of this quadratic equation are a = 1, b = 10, and c = 25. This simplified form allows us to apply standard methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. The substitution technique is a powerful tool in algebra because it allows us to simplify complex expressions by replacing repeating terms with single variables. This not only makes the equations easier to solve but also provides a clearer understanding of the underlying structure. Once we have solved for 'u', we can substitute back the original expression (5x - 4) to find the values of 'x' that satisfy the original equation. This two-step process of substitution and back-substitution is a common strategy for solving equations that initially appear daunting. In the context of this specific problem, the substitution has transformed a complex equation into a familiar quadratic form, setting the stage for the next step: solving the quadratic equation for 'u'. The ability to recognize patterns and apply appropriate substitutions is a valuable skill in algebra and beyond. It demonstrates a deep understanding of mathematical structures and the flexibility to manipulate equations to make them more tractable. By mastering this technique, students and practitioners alike can tackle a wider range of problems with greater confidence and efficiency.
Solving the Quadratic Equation
Now that we have rewritten the equation as u^2 + 10u + 25 = 0, we can solve for 'u'. This quadratic equation is a perfect square trinomial, which means it can be factored easily. A perfect square trinomial is a trinomial that can be expressed as the square of a binomial. In this case, u^2 + 10u + 25 can be factored as (u + 5)^2. So, the equation becomes (u + 5)^2 = 0. To solve for 'u', we take the square root of both sides of the equation. This gives us u + 5 = 0. Subtracting 5 from both sides, we find that u = -5. Since (u + 5)^2 = 0 has a repeated root, there is only one solution for 'u', which is u = -5. This means that the quadratic equation has a single real root. Now that we have found the value of 'u', we can substitute back to find the value of 'x'. Remember that we defined u = (5x - 4). So, we substitute -5 for 'u' in this equation: -5 = 5x - 4. Adding 4 to both sides, we get -1 = 5x. Dividing both sides by 5, we find that x = -1/5. Therefore, the original equation has one real solution, which is x = -1/5. This solution satisfies the original equation, and we have successfully found the real solution using substitution and factoring.
With the equation now in the simplified quadratic form u^2 + 10u + 25 = 0, the next step is to solve for the variable 'u'. This equation is a classic example of a perfect square trinomial. A perfect square trinomial is a quadratic expression that can be factored into the square of a binomial. Recognizing this pattern is crucial for efficient problem-solving. In this case, u^2 + 10u + 25 can be factored as (u + 5)^2. This factorization is based on the principle that (a + b)^2 = a^2 + 2ab + b^2. Comparing this to our equation, we can see that a = u and b = 5, which confirms that it is indeed a perfect square trinomial. Now, our equation becomes (u + 5)^2 = 0. To solve for 'u', we take the square root of both sides of the equation. The square root of (u + 5)^2 is simply (u + 5), and the square root of 0 is 0. Thus, we have u + 5 = 0. This is a simple linear equation that can be easily solved by subtracting 5 from both sides. This yields u = -5. This result indicates that the quadratic equation has a repeated root. A repeated root occurs when the discriminant (b^2 - 4ac) of the quadratic equation is equal to zero. In our case, the discriminant is 10^2 - 4(1)(25) = 100 - 100 = 0, confirming the presence of a repeated root. The fact that we have a single value for 'u' simplifies the next step, which is to substitute back and solve for 'x'. This is a common occurrence when dealing with perfect square trinomials, and it often leads to a more straightforward solution process. The ability to identify and factor perfect square trinomials is a valuable skill in algebra. It not only simplifies the solving process but also demonstrates a strong understanding of algebraic patterns and structures. By mastering this technique, students can approach quadratic equations with greater confidence and efficiency. In the next section, we will use the value of 'u' that we have found to determine the corresponding value(s) of 'x' that satisfy the original equation.
Finding the Solution for x
Having found that u = -5, we now need to substitute back to find the corresponding value(s) of 'x'. Recall that we defined u = (5x - 4). So, we substitute -5 for 'u' in this equation: -5 = 5x - 4. This is a linear equation in 'x', which we can solve using basic algebraic operations. To isolate 'x', we first add 4 to both sides of the equation: -5 + 4 = 5x - 4 + 4, which simplifies to -1 = 5x. Next, we divide both sides of the equation by 5 to solve for 'x': -1/5 = 5x/5, which simplifies to x = -1/5. Therefore, the solution to the original equation is x = -1/5. To verify this solution, we can substitute x = -1/5 back into the original equation: (5(-1/5) - 4)^2 + 10(5(-1/5) - 4) + 25 = 0. Simplifying the expression inside the parentheses, we get (-1 - 4)^2 + 10(-1 - 4) + 25 = 0. This further simplifies to (-5)^2 + 10(-5) + 25 = 0, which becomes 25 - 50 + 25 = 0. The equation holds true, so x = -1/5 is indeed the correct solution. Since the quadratic equation in 'u' had a repeated root, we found only one solution for 'x'. This indicates that the original equation has only one real solution. The process of substituting back and verifying the solution is a crucial step in solving equations. It ensures that the solution we have found is correct and satisfies the original equation. By following this step-by-step process, we can confidently solve complex equations and verify our results.
After successfully determining the value of 'u' to be -5, the subsequent step involves finding the corresponding value(s) of 'x'. This requires substituting the value of 'u' back into the original substitution equation, which we defined as u = (5x - 4). This substitution allows us to transform the equation back into terms of 'x', enabling us to solve for the variable of interest. Substituting u = -5 into the equation u = (5x - 4), we obtain the equation -5 = 5x - 4. This is now a linear equation in 'x', which can be solved using standard algebraic techniques. To isolate 'x', we first add 4 to both sides of the equation. This gives us -5 + 4 = 5x - 4 + 4, which simplifies to -1 = 5x. Next, we divide both sides of the equation by 5 to solve for 'x'. This yields x = -1/5. Therefore, the solution to the original equation is x = -1/5. This is the only real solution to the equation, as the quadratic equation in 'u' had a repeated root. To ensure the accuracy of our solution, it is always a good practice to verify it by substituting it back into the original equation. Substituting x = -1/5 into the equation (5x - 4)^2 + 10(5x - 4) + 25 = 0, we get (5(-1/5) - 4)^2 + 10(5(-1/5) - 4) + 25 = 0. Simplifying the expression inside the parentheses, we have (-1 - 4)^2 + 10(-1 - 4) + 25 = 0. This further simplifies to (-5)^2 + 10(-5) + 25 = 0, which becomes 25 - 50 + 25 = 0. The equation holds true, confirming that x = -1/5 is indeed the correct solution. This process of verification is crucial in mathematics as it helps to identify any potential errors in the solution process. By substituting the solution back into the original equation, we can confirm that it satisfies the equation and that our solution is accurate. In summary, we have successfully solved the original equation by using a substitution to transform it into a quadratic form, solving the quadratic equation for 'u', and then substituting back to find the value of 'x'. This step-by-step approach demonstrates the power of algebraic techniques in solving complex equations.
Conclusion
In this article, we have successfully found the real solution of the equation (5x - 4)^2 + 10(5x - 4) + 25 = 0. We started by recognizing the repeating expression (5x - 4) and making a substitution to simplify the equation into a quadratic form. By letting u = (5x - 4), we transformed the original equation into u^2 + 10u + 25 = 0. This quadratic equation was easily factored as (u + 5)^2 = 0, which gave us the solution u = -5. Substituting back, we found that 5x - 4 = -5, which led to the solution x = -1/5. We verified this solution by substituting it back into the original equation and confirming that it satisfies the equation. This problem demonstrates the power of substitution as a technique for simplifying complex equations. By recognizing patterns and making appropriate substitutions, we can often transform complex equations into simpler forms that we can solve using standard techniques. This approach is widely used in algebra and calculus to solve various types of equations. The ability to recognize patterns and apply appropriate techniques is a key skill in mathematics. By mastering these skills, we can confidently tackle a wide range of problems and gain a deeper understanding of mathematical concepts.
In conclusion, we have successfully navigated the process of finding the real solution to the equation (5x - 4)^2 + 10(5x - 4) + 25 = 0. Our journey began with recognizing the inherent complexity of the equation and the need for a strategic approach to simplification. By identifying the repeating expression (5x - 4), we were able to employ the technique of substitution, a powerful tool in algebra for transforming equations into more manageable forms. The substitution u = (5x - 4) proved to be the key to unlocking the equation's solution. This transformation led us to the quadratic equation u^2 + 10u + 25 = 0, a form that is readily recognizable and solvable. We observed that this quadratic equation is a perfect square trinomial, which allowed us to factor it as (u + 5)^2 = 0. This factorization provided us with the direct solution u = -5. With the value of 'u' in hand, we proceeded to the crucial step of substituting back to find the corresponding value of 'x'. By substituting u = -5 into the equation u = (5x - 4), we obtained the linear equation 5x - 4 = -5. Solving this equation, we arrived at the solution x = -1/5. To ensure the accuracy of our solution, we performed a verification step, substituting x = -1/5 back into the original equation. This substitution confirmed that our solution satisfies the equation, providing us with confidence in our result. Throughout this process, we have highlighted the importance of pattern recognition, strategic substitution, and the application of algebraic techniques. These skills are fundamental to problem-solving in mathematics and beyond. The ability to recognize patterns allows us to identify opportunities for simplification, while strategic substitution enables us to transform complex problems into more manageable forms. By mastering these skills, students and practitioners alike can approach mathematical challenges with greater confidence and efficiency. In summary, the solution to the equation (5x - 4)^2 + 10(5x - 4) + 25 = 0 is x = -1/5. This solution was obtained through a systematic approach involving substitution, factoring, and algebraic manipulation, demonstrating the power of mathematical techniques in solving complex problems.