Solving Equations With Substitution: A Step-by-Step Guide

by Alex Johnson 58 views

Let's dive into solving the equation (x+1)2−4(x+1)+2=0(x+1)^2 - 4(x+1) + 2 = 0 using substitution. This technique can simplify complex equations, making them easier to handle. We'll explore the best choice for our substitution variable, uu, and then walk through the solution process step by step. This method is widely used in mathematics to transform complicated expressions into simpler, more manageable forms. By identifying a repeating pattern within the equation, we can replace it with a single variable, which helps to reduce the complexity and makes the equation easier to solve. Substitution is not only useful for solving algebraic equations but also for simplifying integrals in calculus and solving differential equations. It is a powerful tool that every math student should master. Understanding when and how to use substitution effectively can significantly improve problem-solving skills and efficiency. Moreover, the ability to recognize patterns and apply appropriate substitutions is a valuable skill that extends beyond the realm of mathematics, aiding in various analytical and problem-solving scenarios in other fields as well. This technique highlights the beauty of mathematics in its ability to simplify and elegantly solve problems that initially appear daunting.

Choosing the Right Substitution

The key to effective substitution is choosing the right expression to replace with a new variable. In our equation, (x+1)2−4(x+1)+2=0(x+1)^2 - 4(x+1) + 2 = 0, notice that the term (x+1)(x+1) appears twice. This repetition is a clear indicator that substituting u=x+1u = x+1 would be a smart move. Let's consider why the other options aren't as suitable:

  • If we chose u=xu = x, we'd have to rewrite (x+1)(x+1) in terms of xx, which doesn't simplify the equation directly.
  • If we chose u=(x+1)2u = (x+1)^2, the equation would still contain (x+1)(x+1), making it less simplified than if we chose u=x+1u = x+1.
  • If we chose u=4(x+1)u = 4(x+1), while it does appear in the equation, substituting this would still leave us with a more complex expression involving (x+1)2(x+1)^2. Choosing u=x+1u = x+1 allows for a direct and significant simplification. The goal of substitution is to reduce the complexity of the equation, and selecting the repeating term is often the most effective strategy. By doing so, we transform the original equation into a simpler quadratic form that is much easier to solve. This approach not only simplifies the algebraic manipulation but also provides a clearer path to finding the solutions. Furthermore, the appropriate choice of substitution can often reveal underlying structures and symmetries within the equation, leading to a deeper understanding of the problem and its solution. Therefore, recognizing the repeating terms and making a suitable substitution is a crucial step in solving complex equations efficiently and accurately.

Therefore, the correct choice is A. u=x+1u = x+1.

Solving the Substituted Equation

Now that we've chosen u=x+1u = x+1, let's substitute it into the original equation:

(x+1)2−4(x+1)+2=0(x+1)^2 - 4(x+1) + 2 = 0 becomes u2−4u+2=0u^2 - 4u + 2 = 0.

This is a quadratic equation in terms of uu. We can solve it using the quadratic formula:

u=−b±b2−4ac2au = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In our case, a=1a = 1, b=−4b = -4, and c=2c = 2. Plugging these values into the quadratic formula, we get:

u=−(−4)±(−4)2−4(1)(2)2(1)u = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}

u=4±16−82u = \frac{4 \pm \sqrt{16 - 8}}{2}

u=4±82u = \frac{4 \pm \sqrt{8}}{2}

u=4±222u = \frac{4 \pm 2\sqrt{2}}{2}

u=2±2u = 2 \pm \sqrt{2}

So, we have two possible values for uu: u=2+2u = 2 + \sqrt{2} and u=2−2u = 2 - \sqrt{2}. The quadratic formula is a powerful tool for solving equations of the form ax2+bx+c=0ax^2 + bx + c = 0. It provides a direct method to find the roots, regardless of whether the equation can be easily factored. Understanding and applying the quadratic formula is essential for solving a wide range of mathematical problems. In this case, using the quadratic formula allowed us to find the values of uu without resorting to more complicated methods such as completing the square. The formula is derived from the process of completing the square, and it offers a streamlined approach to finding the solutions. By substituting the coefficients aa, bb, and cc into the formula, we can quickly determine the values of the unknown variable, uu. This highlights the efficiency and utility of the quadratic formula in solving quadratic equations. Mastering this formula is a fundamental step in developing strong algebraic skills.

Finding the Values of x

Since u=x+1u = x + 1, we can now solve for xx by substituting back the values of uu we found:

  1. For u=2+2u = 2 + \sqrt{2}:

    x+1=2+2x + 1 = 2 + \sqrt{2}

    x=2+2−1x = 2 + \sqrt{2} - 1

    x=1+2x = 1 + \sqrt{2}

  2. For u=2−2u = 2 - \sqrt{2}:

    x+1=2−2x + 1 = 2 - \sqrt{2}

    x=2−2−1x = 2 - \sqrt{2} - 1

    x=1−2x = 1 - \sqrt{2}

Thus, the solutions for xx are x=1+2x = 1 + \sqrt{2} and x=1−2x = 1 - \sqrt{2}. The process of substituting back the original variable is a crucial step in solving equations using substitution. It allows us to express the solutions in terms of the original variables and provides a complete answer to the problem. By substituting u=x+1u = x + 1 back into the equations, we were able to find the corresponding values of xx that satisfy the original equation. This step ensures that the solutions are meaningful within the context of the original problem. Moreover, it demonstrates the importance of carefully tracking the substitutions and ensuring that the final answers are expressed in the appropriate variables. This technique is widely used in various areas of mathematics and engineering, where complex problems are often simplified using substitutions. The ability to effectively substitute back and forth between variables is a valuable skill that enhances problem-solving capabilities and ensures accurate results.

Conclusion

By using substitution, we transformed a seemingly complex equation into a simpler quadratic equation, solved for the substituted variable, and then found the values of the original variable. This method demonstrates the power and elegance of substitution in simplifying mathematical problems.

In summary, the solutions to the equation (x+1)2−4(x+1)+2=0(x+1)^2 - 4(x+1) + 2 = 0 are x=1+2x = 1 + \sqrt{2} and x=1−2x = 1 - \sqrt{2}. Through the process of substitution, we were able to reduce the equation to a manageable quadratic form, apply the quadratic formula, and ultimately find the values of xx. This approach highlights the importance of identifying suitable substitutions and utilizing algebraic techniques to solve complex equations. The ability to simplify and solve equations effectively is a fundamental skill in mathematics and engineering. Mastering this skill can lead to a deeper understanding of mathematical concepts and improved problem-solving abilities. Furthermore, the technique of substitution is not only useful for solving algebraic equations but also for simplifying integrals in calculus and solving differential equations. Therefore, understanding and practicing substitution is essential for developing strong mathematical proficiency. The elegance of this method lies in its ability to transform a seemingly daunting problem into a series of simpler, more manageable steps, ultimately leading to a clear and concise solution.

For further reading on algebraic techniques, you might find valuable information on websites like Khan Academy Algebra.