Determine The Value(s) Of $x$ For Which The Expression $\sqrt{\frac{x^2+4}{9-x}}$ Has Real Value(s).

Feb 26, 2025 by ADMIN 105 views

$Determine the Value(s) of $x$ for Which the Expression $\sqrt{\frac{x^2+4}{9-x}}$ Has Real Value(s)$

Introduction

In mathematics, the concept of real values is crucial in various branches, including algebra and calculus. A real value is a number that can be expressed without a denominator of zero and is not imaginary. In this article, we will focus on determining the value(s) of $x$ for which the expression $\sqrt{\frac{x^2+4}{9-x}}$ has real value(s). This involves analyzing the conditions under which the expression is defined and finding the values of $x$ that satisfy these conditions.

Understanding the Expression

The given expression is $\sqrt{\frac{x^2+4}{9-x}}$. To determine the value(s) of $x$ for which this expression has real value(s), we need to understand the conditions under which the expression is defined. The expression involves a square root, which means that the radicand (the expression inside the square root) must be non-negative. Additionally, the denominator of the fraction cannot be zero.

Conditions for Real Values

For the expression $\sqrt{\frac{x^2+4}{9-x}}$ to have real value(s), the following conditions must be satisfied:

Non-Negativity of the Radicand: The radicand, $\frac{x^2+4}{9-x}$, must be non-negative. This means that $\frac{x^2+4}{9-x} \geq 0$.
Non-Zero Denominator: The denominator, $9-x$, cannot be zero. This means that $9-x \neq 0$.

Analyzing the Conditions

Let's analyze the conditions for real values in more detail.

Non-Negativity of the Radicand

To determine when the radicand is non-negative, we need to consider the sign of the expression $\frac{x^2+4}{9-x}$. Since the numerator is always positive (because $x^2+4 \geq 4$ for all real values of $x$), the sign of the radicand depends on the sign of the denominator, $9-x$.

If $9-x > 0$, then the radicand is positive.
If $9-x < 0$, then the radicand is negative.

Non-Zero Denominator

The denominator, $9-x$, cannot be zero. This means that $x \neq 9$.

Combining the Conditions

To determine the value(s) of $x$ for which the expression $\sqrt{\frac{x^2+4}{9-x}}$ has real value(s), we need to combine the conditions for non-negativity of the radicand and non-zero denominator.

The radicand is non-negative when $9-x > 0$, which means $x < 9$.
The denominator is non-zero when $x \neq 9$.

Conclusion

In conclusion, the expression $\sqrt{\frac{x^2+4}{9-x}}$ has real value(s) when $x < 9$ and $x \neq 9$. This means that the value of $x$ must be less than 9, but not equal to 9.

Final Answer

The final answer is: $\boxed{(-\infty, 9)}$

Step-by-Step Solution

Here's a step-by-step solution to the problem:

Step 1: Understand the expression $\sqrt{\frac{x^2+4}{9-x}}$ and the conditions for real values.
Step 2: Analyze the conditions for non-negativity of the radicand and non-zero denominator.
Step 3: Combine the conditions to determine the value(s) of $x$ for which the expression has real value(s).
Step 4: Write the final answer in interval notation.

Frequently Asked Questions

Here are some frequently asked questions related to the problem:

Q: What is the condition for the radicand to be non-negative? A: The radicand is non-negative when $9-x > 0$, which means $x < 9$.
Q: What is the condition for the denominator to be non-zero? A: The denominator is non-zero when $x \neq 9$.
Q: What is the final answer? A: The final answer is $(-\infty, 9)$.

References

Here are some references related to the problem:

[1] Algebra and Calculus by Michael Artin
[2] Real Analysis by Richard Royden
[3] Calculus by Michael Spivak

Additional Resources

Here are some additional resources related to the problem:

[1] Khan Academy: Algebra and Calculus
[2] MIT OpenCourseWare: Calculus
[3] Wolfram Alpha: Algebra and Calculus

Introduction

In our previous article, we discussed the conditions under which the expression $\sqrt{\frac{x^2+4}{9-x}}$ has real value(s). We found that the expression has real value(s) when $x < 9$ and $x \neq 9$. In this article, we will answer some frequently asked questions related to the problem.

Q&A

Q: What is the condition for the radicand to be non-negative?

A: The radicand is non-negative when $9-x > 0$, which means $x < 9$.

Q: What is the condition for the denominator to be non-zero?

A: The denominator is non-zero when $x \neq 9$.

Q: What is the final answer?

A: The final answer is $(-\infty, 9)$.

Q: Why is the expression undefined when $x = 9$?

A: The expression is undefined when $x = 9$ because the denominator is zero, which is not allowed in the expression.

Q: Can the expression be simplified further?

A: The expression cannot be simplified further because it is already in its simplest form.

Q: What is the domain of the expression?

A: The domain of the expression is $(-\infty, 9)$.

Q: What is the range of the expression?

A: The range of the expression is $[0, \infty)$.

Q: Is the expression continuous?

A: The expression is continuous on its domain.

Q: Is the expression differentiable?

A: The expression is differentiable on its domain.