What Is The Solution Of X + 12 = X \sqrt{x+12}=x X + 12 ​ = X ?A. X = − 3 X=-3 X = − 3 B. X = 4 X=4 X = 4 C. X = − 3 X=-3 X = − 3 Or X = 4 X=4 X = 4 D. No Solution

$$

Introduction

Solving equations involving square roots can be a challenging task, especially when the variable is inside the square root. In this article, we will explore the solution to the equation $\sqrt{x+12}=x$. This equation involves a square root and a variable, making it a classic example of a radical equation.

Understanding the Equation

The given equation is $\sqrt{x+12}=x$. To solve this equation, we need to isolate the variable $x$. The first step is to square both sides of the equation to eliminate the square root. This will give us a new equation that we can solve for $x$.

Squaring Both Sides

When we square both sides of the equation, we get:

$$\left(\sqrt{x+12}\right)^2=x^2$$

This simplifies to:

$$x+12=x^2$$

Rearranging the Equation

Now, we need to rearrange the equation to get all the terms on one side. We can do this by subtracting $x$ from both sides and subtracting $12$ from both sides:

$$x^2-x-12=0$$

Factoring the Quadratic Equation

The equation $x^2-x-12=0$ is a quadratic equation. We can factor this equation to find the values of $x$ that satisfy the equation. The factored form of the equation is:

$$(x-4)(x+3)=0$$

Solving for $x$

Now, we can solve for $x$ by setting each factor equal to zero:

$$x-4=0 \quad \text{or} \quad x+3=0$$

Solving for $x$ in each equation, we get:

$$x=4 \quad \text{or} \quad x=-3$$

Checking the Solutions

Before we conclude that $x=4$ and $x=-3$ are the solutions to the equation, we need to check if these values satisfy the original equation. We can do this by plugging each value back into the original equation:

$$\sqrt{4+12}=4 \quad \text{and} \quad \sqrt{-3+12}=-3$$

Simplifying each expression, we get:

$$\sqrt{16}=4 \quad \text{and} \quad \sqrt{9}=-3$$

Since $\sqrt{16}=4$ is true, but $\sqrt{9}=-3$ is not true, we can conclude that only $x=4$ is a valid solution to the equation.

Conclusion

In this article, we solved the equation $\sqrt{x+12}=x$ by squaring both sides, rearranging the equation, factoring the quadratic equation, and solving for $x$. We found that the only valid solution to the equation is $x=4$. Therefore, the correct answer is:

The final answer is B. $x=4$

Discussion

The equation $\sqrt{x+12}=x$ is a classic example of a radical equation. Solving this equation involves squaring both sides, rearranging the equation, factoring the quadratic equation, and solving for $x$. In this article, we showed that the only valid solution to the equation is $x=4$. If you have any questions or comments about this article, please feel free to ask.

Related Articles

• Solving Equations with Square Roots
• Radical Equations
• Quadratic Equations

References

• [1] "Algebra and Trigonometry" by Michael Sullivan
• [2] "College Algebra" by James Stewart
• [3] "Mathematics for the Nonmathematician" by Morris Kline
  

$$

Introduction

In our previous article, we solved the equation $\sqrt{x+12}=x$ and found that the only valid solution to the equation is $x=4$. However, we received many questions and comments from readers who were unsure about the steps involved in solving the equation. In this article, we will address some of the most frequently asked questions about solving the equation $\sqrt{x+12}=x$.

Q: What is the first step in solving the equation $\sqrt{x+12}=x$?

A: The first step in solving the equation $\sqrt{x+12}=x$ is to square both sides of the equation. This will eliminate the square root and give us a new equation that we can solve for $x$.

Q: Why do we need to square both sides of the equation?

A: We need to square both sides of the equation to eliminate the square root. This is because the square root of a number is equal to the number raised to the power of 1/2. By squaring both sides of the equation, we are essentially raising both sides to the power of 2, which will eliminate the square root.

Q: How do we rearrange the equation after squaring both sides?

A: After squaring both sides of the equation, we need to rearrange the equation to get all the terms on one side. We can do this by subtracting $x$ from both sides and subtracting $12$ from both sides. This will give us a new equation that we can solve for $x$.

Q: What is the next step after rearranging the equation?

A: After rearranging the equation, we need to factor the quadratic equation. This will give us two possible solutions for $x$. We can then check each solution to see if it satisfies the original equation.

Q: Why do we need to check each solution?

A: We need to check each solution to make sure that it satisfies the original equation. This is because the equation $\sqrt{x+12}=x$ is a radical equation, and the solutions to the equation may not be what we expect.

Q: What is the final answer to the equation $\sqrt{x+12}=x$?

A: The final answer to the equation $\sqrt{x+12}=x$ is $x=4$. This is the only valid solution to the equation.

Q: Can you explain why $x=-3$ is not a valid solution?

A: $x=-3$ is not a valid solution to the equation $\sqrt{x+12}=x$ because it does not satisfy the original equation. When we plug $x=-3$ back into the original equation, we get $\sqrt{-3+12}=-3$, which is not true.

Q: What is the most important thing to remember when solving radical equations?

A: The most important thing to remember when solving radical equations is to check each solution to make sure that it satisfies the original equation. This will ensure that we get the correct solution to the equation.

Q: Can you provide more examples of radical equations?

A: Yes, here are a few more examples of radical equations:

• $\sqrt{x-5}=x+2$
• $\sqrt{x+1}=x-3$
• $\sqrt{x-2}=x+1$

These equations can be solved using the same steps that we used to solve the equation $\sqrt{x+12}=x$.

Conclusion

In this article, we answered some of the most frequently asked questions about solving the equation $\sqrt{x+12}=x$. We hope that this article has been helpful in clarifying the steps involved in solving radical equations. If you have any more questions or comments, please feel free to ask.

Related Articles

• Solving Equations with Square Roots
• Radical Equations
• Quadratic Equations

References

• [1] "Algebra and Trigonometry" by Michael Sullivan
• [2] "College Algebra" by James Stewart
• [3] "Mathematics for the Nonmathematician" by Morris Kline