Solve: $x=4+(4x-4)^{\frac{1}{2}}$A. $x=2$ B. $x=10$ C. $x=2$ Or $x=10$ D. No Real Solution

Solving the Equation: $x=4+(4x-4)^{\frac{1}{2}}$

In this article, we will delve into the world of mathematics and solve a seemingly complex equation. The equation in question is $x=4+(4x-4)^{\frac{1}{2}}$. We will break down the solution step by step, using algebraic manipulations and mathematical reasoning to arrive at the final answer.

Understanding the Equation

The given equation is $x=4+(4x-4)^{\frac{1}{2}}$. To begin solving this equation, we need to understand the structure of the equation and identify the key components. The equation consists of a square root term, $(4x-4)^{\frac{1}{2}}$, which is added to the constant term, 4. The variable $x$ appears in both the square root term and the constant term.

Squaring Both Sides

One approach to solving this equation is to square both sides of the equation. This will eliminate the square root term and allow us to work with a simpler equation. Squaring both sides of the equation gives us:

$x^2 = 16 + 8x + (4x-4)$

Simplifying the Equation

We can simplify the equation by combining like terms. The equation becomes:

$x^2 = 16 + 8x + 4x - 4$

$x^2 = 16 + 12x - 4$

$x^2 = 12x + 12$

Rearranging the Equation

To make the equation more manageable, we can rearrange it by subtracting $12x$ from both sides. This gives us:

$x^2 - 12x = 12$

Factoring the Equation

The equation can be factored by grouping. We can factor out a common term, $x$, from the left-hand side of the equation:

$x(x - 12) = 12$

Solving for $x$

We can now solve for $x$ by setting each factor equal to zero. This gives us two possible solutions:

$x = 0$ or $x - 12 = 0$

$x = 12$

Checking the Solutions

We need to check each solution to ensure that it satisfies the original equation. Substituting $x = 0$ into the original equation gives us:

$0 = 4 + (4(0) - 4)^{\frac{1}{2}}$

$0 = 4 + (-4)^{\frac{1}{2}}$

$0 = 4 - 2$

$0 \neq 2$

This solution does not satisfy the original equation.

Substituting $x = 12$ into the original equation gives us:

$12 = 4 + (4(12) - 4)^{\frac{1}{2}}$

$12 = 4 + (48 - 4)^{\frac{1}{2}}$

$12 = 4 + (44)^{\frac{1}{2}}$

$12 = 4 + 2\sqrt{11}$

$12 \neq 4 + 2\sqrt{11}$

This solution does not satisfy the original equation.

We have attempted to solve the equation $x=4+(4x-4)^{\frac{1}{2}}$ using algebraic manipulations and mathematical reasoning. However, we have found that neither of the solutions, $x = 0$ or $x = 12$, satisfies the original equation. This suggests that the equation may not have any real solutions.

Final Answer

The final answer is: $\boxed{D}$
Solving the Equation: $x=4+(4x-4)^{\frac{1}{2}}$ - Q&A

In our previous article, we attempted to solve the equation $x=4+(4x-4)^{\frac{1}{2}}$ using algebraic manipulations and mathematical reasoning. However, we found that neither of the solutions, $x = 0$ or $x = 12$, satisfied the original equation. This suggests that the equation may not have any real solutions.

In this article, we will address some of the common questions and concerns that readers may have regarding the solution to this equation.

Q: What is the significance of the square root term in the equation?

A: The square root term, $(4x-4)^{\frac{1}{2}}$, is a key component of the equation. It represents a value that is equal to the square root of the expression $4x-4$. The square root term is added to the constant term, 4, to form the equation.

Q: Why did we square both sides of the equation?

A: Squaring both sides of the equation eliminated the square root term and allowed us to work with a simpler equation. This is a common technique used in algebra to solve equations that involve square roots.

Q: Why did we not find any real solutions to the equation?

A: We did not find any real solutions to the equation because the solutions we obtained, $x = 0$ and $x = 12$, did not satisfy the original equation. This suggests that the equation may not have any real solutions.

Q: What does it mean for an equation to have no real solutions?

A: An equation has no real solutions if there are no values of the variable that satisfy the equation. In other words, there are no values of $x$ that make the equation true.

Q: Can we use other methods to solve the equation?

A: Yes, there are other methods that we can use to solve the equation. For example, we can use numerical methods, such as the Newton-Raphson method, to approximate the solution to the equation.

Q: What are some common mistakes that people make when solving equations with square roots?

A: Some common mistakes that people make when solving equations with square roots include:

• Not squaring both sides of the equation
• Not checking the solutions to ensure that they satisfy the original equation
• Not considering the possibility that the equation may have no real solutions

In this article, we have addressed some of the common questions and concerns that readers may have regarding the solution to the equation $x=4+(4x-4)^{\frac{1}{2}}$. We have discussed the significance of the square root term, the importance of squaring both sides of the equation, and the possibility that the equation may have no real solutions. We hope that this article has been helpful in clarifying some of the concepts and techniques involved in solving equations with square roots.

Final Answer

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