What Are The Solutions Of The Quadratic Equation $(x-8)^2 - 13(x-8) + 30 = 0$?Use $u$ Substitution To Solve.A. $x = -11$ And $ X = − 18 X = -18 X = − 18 [/tex] B. $x = -2$ And $x = 5$ C.

Introduction

Solving quadratic equations is a fundamental concept in algebra, and there are various methods to find the solutions. One of the methods is the substitution method, where we substitute a variable with an expression to simplify the equation. In this article, we will use the $u$ substitution method to solve the quadratic equation $(x-8)^2 - 13(x-8) + 30 = 0$.

Step 1: Understand the equation

The given equation is $(x-8)^2 - 13(x-8) + 30 = 0$. We can see that the equation is in the form of a quadratic equation, where the variable $x$ is squared and multiplied by a constant. The equation also contains a linear term and a constant term.

Step 2: Use $u$ substitution

To simplify the equation, we can use the $u$ substitution method. We will substitute $u = x - 8$, which means $x = u + 8$. By substituting this expression into the equation, we can simplify the equation and make it easier to solve.

Step 3: Substitute $u$ into the equation

Substituting $u = x - 8$ into the equation, we get:

$$(u + 8)^2 - 13(u + 8) + 30 = 0$$

Expanding the squared term, we get:

$$u^2 + 16u + 64 - 13u - 104 + 30 = 0$$

Combining like terms, we get:

$$u^2 + 3u - 30 = 0$$

Step 4: Solve the quadratic equation

Now we have a quadratic equation in the form of $u^2 + 3u - 30 = 0$. We can solve this equation using the quadratic formula or factoring. Let's use factoring to solve the equation.

Step 5: Factor the quadratic equation

The quadratic equation $u^2 + 3u - 30 = 0$ can be factored as:

$$(u + 6)(u - 5) = 0$$

Step 6: Solve for $u$

Now we have two factors: $(u + 6)$ and $(u - 5)$. We can set each factor equal to zero and solve for $u$.

$$u + 6 = 0 \Rightarrow u = -6$$

$$u - 5 = 0 \Rightarrow u = 5$$

Step 7: Substitute back $u$

Now that we have the values of $u$, we can substitute back $u = x - 8$ to find the values of $x$.

For $u = -6$, we have:

$$x - 8 = -6 \Rightarrow x = -6 + 8 = 2$$

For $u = 5$, we have:

$$x - 8 = 5 \Rightarrow x = 5 + 8 = 13$$

Step 8: Check the solutions

We have found two solutions: $x = 2$ and $x = 13$. We can check these solutions by substituting them back into the original equation.

For $x = 2$, we have:

$$(2 - 8)^2 - 13(2 - 8) + 30 = (-6)^2 - 13(-6) + 30 = 36 + 78 + 30 = 144$$

For $x = 13$, we have:

$$(13 - 8)^2 - 13(13 - 8) + 30 = (5)^2 - 13(5) + 30 = 25 - 65 + 30 = -10$$

Since $x = 13$ does not satisfy the equation, we can conclude that the only solution is $x = 2$.

Conclusion

In this article, we used the $u$ substitution method to solve the quadratic equation $(x-8)^2 - 13(x-8) + 30 = 0$. We found that the only solution is $x = 2$. This method is useful for simplifying quadratic equations and making them easier to solve.

Final Answer

The final answer is: $\boxed{2}$

Introduction

Solving quadratic equations can be a challenging task, but with the right approach, it can be made easier. In our previous article, we used the $u$ substitution method to solve the quadratic equation $(x-8)^2 - 13(x-8) + 30 = 0$. In this article, we will provide a Q&A guide to help you understand the concept better.

Q: What is a quadratic equation?

A: A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. It is typically written in the form of $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants.

Q: What is the $u$ substitution method?

A: The $u$ substitution method is a technique used to simplify quadratic equations by substituting a variable with an expression. In this method, we substitute $u = x - a$, where $a$ is a constant, and then simplify the equation.

Q: How do I use the $u$ substitution method?

A: To use the $u$ substitution method, follow these steps:

1. Identify the constant $a$ in the equation.
2. Substitute $u = x - a$ into the equation.
3. Simplify the equation by combining like terms.
4. Factor the equation, if possible.
5. Solve for $u$.
6. Substitute back $u = x - a$ to find the values of $x$.

Q: What are the common mistakes to avoid when using the $u$ substitution method?

A: Some common mistakes to avoid when using the $u$ substitution method include:

• Not identifying the constant $a$ correctly.
• Not simplifying the equation properly.
• Not factoring the equation correctly.
• Not substituting back $u = x - a$ correctly.

Q: Can I use the $u$ substitution method for all quadratic equations?

A: No, the $u$ substitution method is not suitable for all quadratic equations. It is best used for equations that have a constant term that can be factored out.

Q: What are the advantages of using the $u$ substitution method?

A: The advantages of using the $u$ substitution method include:

• Simplifying the equation by reducing the degree of the polynomial.
• Making it easier to factor the equation.
• Reducing the number of solutions.

Q: What are the disadvantages of using the $u$ substitution method?

A: The disadvantages of using the $u$ substitution method include:

• It can be time-consuming to identify the constant $a$ and substitute $u = x - a$.
• It may not be suitable for all quadratic equations.
• It requires careful simplification and factoring of the equation.

Q: Can I use other methods to solve quadratic equations?

A: Yes, there are other methods to solve quadratic equations, including:

• Factoring
• Quadratic formula
• Graphing
• Completing the square

Q: Which method is the best for solving quadratic equations?

A: The best method for solving quadratic equations depends on the specific equation and the individual's preference. The $u$ substitution method is a useful technique, but it may not be the best method for all equations.

Conclusion

In this Q&A guide, we have provided an overview of the $u$ substitution method and its applications. We have also discussed common mistakes to avoid and the advantages and disadvantages of using this method. Whether you are a student or a teacher, this guide will help you understand the concept better and make it easier to solve quadratic equations.

Final Answer

The final answer is: $\boxed{2}$