Solving (x-8)^2 - 13(x-8) + 30 = 0 Using U-Substitution

In mathematics, quadratic equations play a pivotal role, appearing in various fields from physics to engineering. Mastering the techniques to solve these equations is crucial for anyone delving into these disciplines. One such powerful technique is u-substitution, which simplifies complex quadratic equations into a more manageable form. This article aims to provide a comprehensive guide on solving quadratic equations using u-substitution, with a focus on a specific example to illustrate the process step-by-step.

Understanding Quadratic Equations

Before diving into the intricacies of u-substitution, it's essential to grasp the fundamental concept of quadratic equations. A quadratic equation is a polynomial equation of the second degree, generally expressed in the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. The solutions to this equation, also known as roots or zeros, represent the values of x that satisfy the equation. These roots can be real or complex numbers.

There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. However, when dealing with equations that have a more complex structure, u-substitution offers an elegant and efficient approach. This method involves introducing a new variable, u, to replace a recurring expression within the equation, thereby simplifying the equation and making it easier to solve. Once the values of u are found, they are then substituted back to determine the values of the original variable, x.

The Power of U-Substitution

U-substitution is a technique that simplifies complex equations by replacing a recurring expression with a single variable, making the equation easier to solve. This method is particularly useful when dealing with quadratic equations that have a composite expression within them. By substituting u for this expression, the equation transforms into a simpler quadratic form, which can then be solved using standard methods such as factoring or the quadratic formula. The solutions for u are then used to find the solutions for the original variable, x. This approach not only simplifies the algebraic manipulations but also provides a clearer understanding of the equation's structure and solutions.

The beauty of u-substitution lies in its ability to transform a seemingly complicated equation into a familiar form. This transformation not only simplifies the solving process but also enhances our understanding of the underlying mathematical structure. By recognizing patterns and recurring expressions, we can effectively apply u-substitution to a wide range of equations, making it a valuable tool in our problem-solving arsenal. Moreover, the technique is not limited to quadratic equations; it can be extended to other types of equations, including those involving trigonometric, exponential, and logarithmic functions. This versatility makes u-substitution a fundamental concept in algebra and calculus.

Solving (x-8)² - 13(x-8) + 30 = 0 Using U-Substitution

Let's consider the quadratic equation (x-8)² - 13(x-8) + 30 = 0. This equation, at first glance, might appear daunting. However, by employing u-substitution, we can transform it into a more manageable form. The key observation here is the recurring expression (x-8). We can simplify the equation by substituting this expression with a new variable, u. This substitution will reveal the underlying quadratic structure and allow us to solve for x more easily.

Step 1: Introduce the Substitution

We begin by letting u = x - 8. This substitution is the cornerstone of the u-substitution method. By replacing the complex expression (x-8) with the single variable u, we simplify the equation and make it easier to work with. This step is crucial for transforming the equation into a standard quadratic form, which we can then solve using familiar techniques. The choice of u is often guided by the structure of the equation, with the aim of identifying a recurring expression that can be replaced to simplify the equation.

Step 2: Rewrite the Equation in Terms of u

Substituting u = x - 8 into the original equation, we get u² - 13u + 30 = 0. This new equation is a standard quadratic equation in terms of u. Notice how the substitution has transformed the original equation, which had a composite expression, into a simple quadratic equation that we can readily solve. This step highlights the power of u-substitution in simplifying complex equations and making them more accessible.

Step 3: Solve the Quadratic Equation for u

Now we need to solve the quadratic equation u² - 13u + 30 = 0 for u. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, the equation can be easily factored. Factoring involves expressing the quadratic expression as a product of two linear factors. This method is particularly efficient when the coefficients of the quadratic equation are integers and the roots are rational numbers.

By factoring, we find that u² - 13u + 30 = (u - 3)(u - 10) = 0. This factorization reveals the roots of the equation in terms of u. The roots are the values of u that make the equation equal to zero. Setting each factor equal to zero, we get u - 3 = 0 and u - 10 = 0. Solving these linear equations, we find the solutions for u: u = 3 and u = 10. These values of u are the roots of the transformed quadratic equation and represent the intermediate solutions in our quest to find the values of x.

Step 4: Substitute Back to Find x

Now that we have the values of u, we need to substitute back to find the corresponding values of x. Recall that we defined u = x - 8. This relationship allows us to express x in terms of u and find the solutions for x. Substituting the values of u we found in the previous step, we can determine the values of x that satisfy the original equation. This step is crucial for obtaining the final solutions in terms of the original variable.

Substituting u = 3, we get 3 = x - 8. Solving for x, we find x = 11. Similarly, substituting u = 10, we get 10 = x - 8. Solving for x, we find x = 18. Therefore, the solutions to the original quadratic equation (x-8)² - 13(x-8) + 30 = 0 are x = 11 and x = 18. These values of x are the roots of the equation and represent the points where the quadratic function intersects the x-axis.

Verifying the Solutions

To ensure the accuracy of our solutions, it's always a good practice to verify them by substituting them back into the original equation. This step helps us to catch any potential errors in our calculations and confirms that the solutions we obtained indeed satisfy the equation. Substituting the values of x back into the original equation, we can check if the equation holds true. This verification step is an essential part of the problem-solving process and provides us with confidence in our solutions.

Substituting x = 11 into the original equation (x-8)² - 13(x-8) + 30 = 0, we get (11-8)² - 13(11-8) + 30 = 3² - 13(3) + 30 = 9 - 39 + 30 = 0. This confirms that x = 11 is a valid solution. Similarly, substituting x = 18 into the original equation, we get (18-8)² - 13(18-8) + 30 = 10² - 13(10) + 30 = 100 - 130 + 30 = 0. This confirms that x = 18 is also a valid solution. Therefore, we have verified that the solutions x = 11 and x = 18 are correct.

Conclusion

In conclusion, the solutions to the quadratic equation (x-8)² - 13(x-8) + 30 = 0 are x = 11 and x = 18. This example illustrates the power and elegance of u-substitution in solving complex quadratic equations. By recognizing recurring expressions and making appropriate substitutions, we can transform complicated equations into simpler forms that are easier to solve. This technique is a valuable tool in the arsenal of any mathematician or problem-solver, and it can be applied to a wide range of equations beyond quadratics.

Understanding and mastering u-substitution is not just about finding solutions to equations; it's about developing a deeper understanding of mathematical structures and problem-solving strategies. By practicing this technique, we can enhance our algebraic skills and become more confident in tackling complex mathematical challenges. The ability to recognize patterns, make substitutions, and simplify equations is a fundamental skill that will serve us well in various mathematical and scientific endeavors. So, embrace the power of u-substitution and unlock new possibilities in your mathematical journey.