https://file.aiquickdraw.com/mathbot/file/172607535910896bdhi1c.jpg

To solve the problem, we need to find the value of \(x\) such that the expression:

\[
\sqrt{x^2 - 9}
\]

is a real and irrational number.

### Step-by-Step Solution:

1. **Expression Analysis**: The square root function \(\sqrt{a}\) is defined and real when \(a \geq 0\). Thus, for \(\sqrt{x^2 - 9}\) to be real, the expression inside the square root, \(x^2 - 9\), must be non-negative:

   \[
   x^2 - 9 \geq 0
   \]

2. **Solving the Inequality**: To find the range of \(x\):

   \[
   x^2 \geq 9
   \]

   Taking the square root on both sides gives:

   \[
   x \geq 3 \quad \text{or} \quad x \leq -3
   \]

3. **Determining When the Expression is Irrational**: For \(\sqrt{x^2 - 9}\) to be irrational, \(x^2 - 9\) must not be a perfect square. 

   Let's check the options:

   - **Option 1: \(x = 5\)**  
     \[
     x^2 - 9 = 5^2 - 9 = 25 - 9 = 16 \quad (\text{perfect square})
     \]
     Result: Real but rational.

   - **Option 2: \(x = 0\)**  
     \[
     x^2 - 9 = 0^2 - 9 = -9 \quad (\text{not real})
     \]
     Result: Not real.

   - **Option 3: \(x = -3\)**  
     \[
     x^2 - 9 = (-3)^2 - 9 = 9 - 9 = 0 \quad (\text{real but rational})
     \]
     Result: Real but rational.

   - **Option 4: \(x = 4\)**  
     \[
     x^2 - 9 = 4^2 - 9 = 16 - 9 = 7 \quad (\text{not a perfect square})
     \]
     Result: Real and irrational.

### Conclusion:

The correct answer is **Option 4: \(x = 4\)** because \(\sqrt{7}\) is a real and irrational number.

---

Would you like further details or have any questions?

### Related Questions:

1. What is the difference between rational and irrational numbers?
2. How do you determine if a square root is irrational?
3. What are some examples of irrational numbers?
4. Can negative numbers inside a square root ever be real?
5. How do you solve inequalities involving quadratic expressions?

### Tip:

Always remember that the square root of a non-perfect square integer is irrational!

The value of \(\sqrt{x^2 - 9}\) is a real and irrational number when x is equal to: 1) 5 2) 0 3) -3 4) 4

Solve the expression \(\sqrt{x^2 - 9}\) to determine when the result is a real and irrational number. This algebraic problem explores square roots, quadratic equations, and the concept of irrational numbers, suitable for students in Grades 7-9.

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