Solving (x+4)^2 - 3(x+4) - 3 = 0 With Substitution A Step-by-Step Guide

$$

In the realm of algebra, solving quadratic equations is a fundamental skill. While some quadratic equations can be solved by factoring or using the quadratic formula directly, others benefit from a clever technique called substitution. Substitution simplifies the equation by introducing a new variable, making it easier to manipulate and solve. This article will delve into the process of solving the equation $(x+4)^2 - 3(x+4) - 3 = 0$ using substitution, providing a step-by-step guide and explaining the underlying concepts. We will specifically focus on how to choose the correct substitution, solve the resulting equation, and then revert back to the original variable to find the solution.

Understanding Substitution in Quadratic Equations

Substitution is a powerful algebraic technique that simplifies complex equations by replacing a complicated expression with a single variable. This often transforms the equation into a more manageable form, such as a standard quadratic equation. The key idea is to identify a recurring expression within the equation and temporarily replace it with a new variable. Once the simplified equation is solved for the new variable, we substitute back the original expression to find the solution for the original variable.

In the given equation, $(x+4)^2 - 3(x+4) - 3 = 0$, the expression $(x+4)$ appears twice. This repetition suggests that substitution is a suitable approach. By letting $u = (x+4)$, we can transform the equation into a simpler quadratic equation in terms of $u$. This makes the equation easier to solve, and then we can substitute back to find the values of $x$. Choosing the correct expression for substitution is crucial for simplifying the equation effectively. A poor choice of substitution might not lead to a simplification and could even make the equation more complex. Therefore, careful observation of the equation's structure is essential for successful substitution.

Step-by-Step Solution Using Substitution

To effectively solve $(x+4)^2 - 3(x+4) - 3 = 0$, let's break down the solution process into manageable steps, ensuring clarity and precision in each stage.

1. Identifying the Appropriate Substitution

As highlighted earlier, the expression $(x+4)$ appears repeatedly in the equation. This makes it an ideal candidate for substitution. Let's define a new variable, $u$, such that:

$u = x + 4$

This substitution aims to simplify the equation's structure and make it more amenable to standard algebraic techniques.

2. Substituting and Forming the Quadratic Equation

Now, we replace every instance of $(x+4)$ in the original equation with $u$. This transforms the equation as follows:

$(x+4)^2 - 3(x+4) - 3 = 0$ becomes

$u^2 - 3u - 3 = 0$

The resulting equation, $u^2 - 3u - 3 = 0$, is a standard quadratic equation in the variable $u$. This simplified form is much easier to work with compared to the original equation.

3. Solving the Quadratic Equation for u

The quadratic equation $u^2 - 3u - 3 = 0$ does not factor easily, so we will use the quadratic formula to find the solutions for $u$. The quadratic formula is given by:

u = rac{-b o \pm o sqrt{b^2 - 4ac}}{2a}

where $a$, $b$, and $c$ are the coefficients of the quadratic equation $au^2 + bu + c = 0$. In our case, $a = 1$, $b = -3$, and $c = -3$.

Substituting these values into the quadratic formula, we get:

u = rac{-(-3) o \pm o sqrt{(-3)^2 - 4(1)(-3)}}{2(1)}

u = rac{3 o \pm o sqrt{9 + 12}}{2}

u = rac{3 o \pm o sqrt{21}}{2}

Thus, we have two solutions for $u$:

u_1 = rac{3 + o sqrt{21}}{2}

u_2 = rac{3 - o sqrt{21}}{2}

4. Substituting Back to Solve for x

Now that we have the values of $u$, we need to substitute back $x + 4$ for $u$ to find the solutions for $x$. We have two equations to solve:

1. x + 4 = rac{3 + o sqrt{21}}{2}

2. x + 4 = rac{3 - o sqrt{21}}{2}

Let's solve each equation separately.

For the first equation:

x + 4 = rac{3 + o sqrt{21}}{2}

x = rac{3 + o sqrt{21}}{2} - 4

x = rac{3 + o sqrt{21} - 8}{2}

x = rac{-5 + o sqrt{21}}{2}

For the second equation:

x + 4 = rac{3 - o sqrt{21}}{2}

x = rac{3 - o sqrt{21}}{2} - 4

x = rac{3 - o sqrt{21} - 8}{2}

x = rac{-5 - o sqrt{21}}{2}

5. Final Solutions for x

Therefore, the solutions for the equation $(x+4)^2 - 3(x+4) - 3 = 0$ are:

x_1 = rac{-5 + o sqrt{21}}{2}

x_2 = rac{-5 - o sqrt{21}}{2}

These are the two values of $x$ that satisfy the original equation.

Why Choose Substitution? Advantages and Applications

Substitution isn't just a trick; it's a strategic tool that simplifies equations by replacing complex expressions with single variables. This method can transform complicated equations into familiar forms, making them easier to solve. In the case of $(x+4)^2 - 3(x+4) - 3 = 0$, substituting $u = (x+4)$ converted the equation into a standard quadratic form, which we could then solve using the quadratic formula. This approach highlights the power of substitution in simplifying algebraic problems.

One of the main advantages of substitution is its ability to make equations more manageable. By reducing the complexity of the equation, we minimize the chances of making errors in algebraic manipulations. This is especially useful in exams or situations where precision is crucial. Substitution also enhances conceptual clarity. When we replace a complex term with a single variable, we can focus on the underlying structure of the equation without being distracted by intricate details. This allows for a more intuitive understanding of the problem and the solution process.

Moreover, substitution is a versatile technique that extends beyond solving quadratic equations. It is widely used in calculus, differential equations, and various branches of mathematics and physics. For example, in calculus, substitution is a fundamental method for evaluating integrals. In differential equations, it is used to transform equations into solvable forms. Understanding substitution provides a solid foundation for tackling more advanced mathematical problems.

Common Mistakes to Avoid When Using Substitution

While substitution is a powerful technique, it's crucial to use it correctly to avoid common pitfalls. One frequent mistake is forgetting to substitute back to the original variable. After solving for the new variable (e.g., u), you must replace it with the original expression (e.g., x + 4) to find the solutions in terms of the original variable. Failing to do so leaves you with the solutions for the substituted variable, not the original one.

Another common mistake is choosing an inappropriate substitution. The goal of substitution is to simplify the equation, so the expression you choose to substitute should appear multiple times in the equation or be part of a more complex term. A poorly chosen substitution can complicate the equation further. Always look for repeating expressions or terms that can be simplified by substitution.

Careless algebraic manipulation after substitution is also a potential source of error. Ensure that you accurately apply the substitution and simplify the resulting equation. Double-check your work at each step to minimize the risk of mistakes. This includes correctly applying the quadratic formula or factoring the simplified equation. A small error in algebra can lead to incorrect solutions, so meticulous attention to detail is essential.

Practice Problems and Further Exploration

To solidify your understanding of substitution, it's essential to practice with various problems. Consider the following examples:

1. Solve $(x-2)^2 + 5(x-2) + 6 = 0$ using substitution.
2. Solve $(2x+1)^2 - 4(2x+1) + 3 = 0$ using substitution.
3. Solve $(x^2 - 1)^2 - 2(x^2 - 1) - 8 = 0$ using substitution.

By working through these problems, you'll gain confidence in identifying appropriate substitutions and solving equations effectively. You can also explore more complex equations where substitution is combined with other techniques, such as factoring or completing the square. Further exploration into advanced applications of substitution in calculus and differential equations can provide a broader perspective on its utility.

In conclusion, mastering substitution is a valuable skill in algebra. It simplifies complex equations, enhances problem-solving abilities, and lays a strong foundation for advanced mathematical concepts. By understanding the principles and practicing regularly, you can confidently apply substitution to a wide range of equations and problems.