Given:${ C = \frac{v}{6} + \sqrt{\frac{3}{v+1}} }$For What Value(s) Of { V $}$ Will { C $}$ Be A Real Number?

Introduction

In mathematics, equations involving square roots and fractions can be complex and challenging to solve. The given equation, $C = \frac{v}{6} + \sqrt{\frac{3}{v+1}}$, involves both a fraction and a square root, making it essential to determine the conditions under which the expression will yield a real number. In this article, we will delve into the world of mathematics and explore the values of $v$ that will make $C$ a real number.

Understanding the Equation

The given equation is $C = \frac{v}{6} + \sqrt{\frac{3}{v+1}}$. To determine the values of $v$ that will make $C$ a real number, we need to analyze the expression inside the square root, $\frac{3}{v+1}$. For the square root to yield a real number, the expression inside it must be non-negative.

Non-Negativity of the Expression Inside the Square Root

For the expression $\frac{3}{v+1}$ to be non-negative, we need to consider two cases:

• Case 1: $v+1 > 0$
• Case 2: $v+1 = 0$
• Case 3: $v+1 < 0$

Let's analyze each case separately.

Case 1: $v+1 > 0$

In this case, $v > -1$. Since $v$ is greater than $-1$, the expression $\frac{3}{v+1}$ will be positive. Therefore, the square root will yield a real number.

Case 2: $v+1 = 0$

In this case, $v = -1$. When $v = -1$, the expression $\frac{3}{v+1}$ becomes $\frac{3}{0}$, which is undefined. Therefore, this case does not yield a real number.

Case 3: $v+1 < 0$

In this case, $v < -1$. Since $v$ is less than $-1$, the expression $\frac{3}{v+1}$ will be negative. Therefore, the square root will not yield a real number.

Conclusion

Based on the analysis of the expression inside the square root, we can conclude that for the given equation to yield a real number, the expression $\frac{3}{v+1}$ must be non-negative. This occurs when $v > -1$. Therefore, the values of $v$ that will make $C$ a real number are $v > -1$.

Final Answer

The final answer is $\boxed{v > -1}$.

Additional Insights

• The given equation involves a square root, which makes it essential to determine the conditions under which the expression will yield a real number.
• The expression inside the square root must be non-negative for the square root to yield a real number.
• The values of $v$ that will make $C$ a real number are $v > -1$.

References

• [1] "Algebra and Trigonometry" by Michael Sullivan
• [2] "Calculus" by James Stewart

Related Topics

• Square roots and fractions
• Non-negativity of expressions
• Real number solutions for equations

Frequently Asked Questions

• Q: What is the given equation? A: The given equation is $C = \frac{v}{6} + \sqrt{\frac{3}{v+1}}$.
• Q: What are the values of $v$ that will make $C$ a real number? A: The values of $v$ that will make $C$ a real number are $v > -1$.
• Q: Why is it essential to determine the conditions under which the expression will yield a real number? A: It is essential to determine the conditions under which the expression will yield a real number because the square root will not yield a real number if the expression inside it is negative.
  Q&A: Real Number Solutions for the Given Equation =====================================================

Introduction

In our previous article, we explored the values of $v$ that will make $C$ a real number in the given equation $C = \frac{v}{6} + \sqrt{\frac{3}{v+1}}$. In this article, we will provide a Q&A section to address some of the most frequently asked questions related to the topic.

Q&A

Q: What is the given equation?

A: The given equation is $C = \frac{v}{6} + \sqrt{\frac{3}{v+1}}$.

Q: What are the values of $v$ that will make $C$ a real number?

A: The values of $v$ that will make $C$ a real number are $v > -1$.

Q: Why is it essential to determine the conditions under which the expression will yield a real number?

A: It is essential to determine the conditions under which the expression will yield a real number because the square root will not yield a real number if the expression inside it is negative.

Q: What happens if $v = -1$?

A: If $v = -1$, the expression $\frac{3}{v+1}$ becomes $\frac{3}{0}$, which is undefined. Therefore, this case does not yield a real number.

Q: Can you provide an example of a value of $v$ that will make $C$ a real number?

A: Yes, a value of $v$ that will make $C$ a real number is $v = 0$. In this case, the expression $\frac{3}{v+1}$ becomes $\frac{3}{1}$, which is positive. Therefore, the square root will yield a real number.

Q: How do you determine the conditions under which the expression will yield a real number?

A: To determine the conditions under which the expression will yield a real number, you need to analyze the expression inside the square root. In this case, the expression is $\frac{3}{v+1}$. You need to consider two cases: $v+1 > 0$ and $v+1 < 0$. If $v+1 > 0$, the expression is positive, and if $v+1 < 0$, the expression is negative.

Q: What is the significance of the expression being non-negative?

A: The expression being non-negative is crucial because the square root will not yield a real number if the expression inside it is negative.

Q: Can you provide a summary of the main points?

A: Yes, the main points are:

• The given equation is $C = \frac{v}{6} + \sqrt{\frac{3}{v+1}}$.
• The values of $v$ that will make $C$ a real number are $v > -1$.
• The expression inside the square root must be non-negative for the square root to yield a real number.
• The expression being non-negative is crucial because the square root will not yield a real number if the expression inside it is negative.

Conclusion

In this Q&A article, we addressed some of the most frequently asked questions related to the topic of real number solutions for the given equation. We hope that this article has provided valuable insights and information to our readers.

Final Answer

The final answer is $\boxed{v > -1}$.

Additional Insights

• The given equation involves a square root, which makes it essential to determine the conditions under which the expression will yield a real number.
• The expression inside the square root must be non-negative for the square root to yield a real number.
• The values of $v$ that will make $C$ a real number are $v > -1$.

References

• [1] "Algebra and Trigonometry" by Michael Sullivan
• [2] "Calculus" by James Stewart

Related Topics

• Square roots and fractions
• Non-negativity of expressions
• Real number solutions for equations

Frequently Asked Questions

• Q: What is the given equation? A: The given equation is $C = \frac{v}{6} + \sqrt{\frac{3}{v+1}}$.
• Q: What are the values of $v$ that will make $C$ a real number? A: The values of $v$ that will make $C$ a real number are $v > -1$.
• Q: Why is it essential to determine the conditions under which the expression will yield a real number? A: It is essential to determine the conditions under which the expression will yield a real number because the square root will not yield a real number if the expression inside it is negative.