Explanation:
Okay, so I need to figure out the real values of x that are not in the domain of the function f(x) = 1 divided by (|x² + 3x - 4| + |x² + 9x + 20|). Hmm, the domain of a function is all the real numbers for which the function is defined. Since this function is a fraction, the denominator can't be zero. So, the values of x that make the denominator zero are not in the domain. Therefore, I need to find all real numbers x where |x² + 3x - 4| + |x² + 9x + 20| = 0. But wait, the sum of two absolute values is zero only if each absolute value is zero individually. Because absolute values are always non-negative, so their sum can only be zero if both terms are zero. So, I need to solve the system:

|x² + 3x - 4| = 0 and |x² + 9x + 20| = 0.

Which implies:

x² + 3x - 4 = 0 and x² + 9x + 20 = 0.

So, solving these two quadratic equations separately and then finding their intersection (since both must be true at the same time) will give me the values of x not in the domain. Let me start with the first equation: x² + 3x - 4 = 0.

To solve this quadratic equation, I can use the quadratic formula. The discriminant D is b² - 4ac = 9 + 16 = 25. So, the roots are (-3 ± √25)/2 = (-3 ± 5)/2. Therefore, the solutions are (-3 + 5)/2 = 2/2 = 1, and (-3 - 5)/2 = -8/2 = -4. So, the first equation has solutions x = 1 and x = -4.

Now, moving on to the second equation: x² + 9x + 20 = 0. Again, using the quadratic formula. The discriminant D is 81 - 80 = 1. So, the roots are (-9 ± √1)/2 = (-9 ± 1)/2. Therefore, the solutions are (-9 + 1)/2 = -8/2 = -4, and (-9 - 1)/2 = -10/2 = -5. So, the second equation has solutions x = -4 and x = -5.

Now, the system requires both equations to be satisfied. So, looking for x values that are in both solutions. The first equation has x = 1 and x = -4. The second equation has x = -4 and x = -5. The only common solution is x = -4. Therefore, x = -4 is the only value where both absolute value expressions are zero. Therefore, the denominator becomes zero at x = -4, so x = -4 is not in the domain of f(x).

Wait, but hold on. Let me check if there are any other possibilities. Because sometimes when dealing with absolute values, there might be cases where the expressions inside are negative, but since we have absolute values, they are always non-negative. However, the sum of two non-negative numbers is zero only if both are zero. Therefore, there are no other possibilities. Therefore, the only x that makes the denominator zero is x = -4. Therefore, the domain of f(x) is all real numbers except x = -4. Therefore, the real value not in the domain is x = -4.

But wait, let me double-check. Let me plug x = -4 into the denominator. Let's compute |(-4)^2 + 3*(-4) - 4| + |(-4)^2 + 9*(-4) + 20|.

First term: 16 - 12 - 4 = 0. Absolute value is 0. Second term: 16 - 36 + 20 = 0. Absolute value is 0. So, 0 + 0 = 0. Therefore, denominator is zero. So, x = -4 is indeed not in the domain.

But let me check if there are other x values where the denominator could be zero. Suppose, for example, that one of the absolute values is positive and the other is negative? Wait, no. Absolute values are always non-negative. So, their sum can't be negative. The only way the sum is zero is if both are zero. Therefore, there are no other solutions. Therefore, only x = -4 is excluded.

But let me think again. Maybe there's a mistake here. Let me check if there are any other points where the expressions inside the absolute values could be zero. For example, maybe if one of them is zero and the other is also zero? But that's exactly what we did. So, the only x that makes both expressions zero is x = -4.

Alternatively, maybe there's a case where one of the absolute values is zero and the other is also zero? Which is the same as above.

Alternatively, maybe there's a case where the expressions inside the absolute values are not zero, but their sum is zero? But since absolute values are non-negative, their sum can't be negative. So, the sum can be zero only if both are zero. Therefore, there are no other solutions.

Therefore, the only value not in the domain is x = -4. Therefore, the answer is x = -4.

But wait, let me check if there are any other points where the denominator could be zero. For example, suppose that one of the absolute values is positive and the other is negative? But absolute values can't be negative. So, even if one is positive and the other is positive, their sum is positive. Therefore, the denominator is always positive except when both are zero. Therefore, only x = -4 is excluded.

But let me check another value. Let me pick x = 1. Let's compute the denominator. First term: |1 + 3 - 4| = |0| = 0. Second term: |1 + 9 + 20| = |30| = 30. Therefore, denominator is 0 + 30 = 30 ≠ 0. Therefore, x = 1 is in the domain. Similarly, x = -5. Let's check. First term: |(-5)^2 + 3*(-5) - 4| = |25 - 15 - 4| = |6| = 6. Second term: |(-5)^2 + 9*(-5) + 20| = |25 - 45 + 20| = |0| = 0. Therefore, denominator is 6 + 0 = 6 ≠ 0. Therefore, x = -5 is in the domain.

Therefore, only x = -4 is excluded. Therefore, the answer is x = -4.

But wait, the question says "real values of x are not in the domain". So, maybe there are more? Let me think again. Suppose that the denominator is zero. So, |A| + |B| = 0. Which implies A = 0 and B = 0. Therefore, solving A = 0 and B = 0. Which gives x = -4. Therefore, only x = -4.

Alternatively, maybe there's a mistake in solving the quadratics? Let me check again.

First quadratic: x² + 3x - 4 = 0. The roots are x = [-3 ± sqrt(9 + 16)] / 2 = [-3 ± 5]/2. So, 2/2 = 1 and -8/2 = -4. Correct.

Second quadratic: x² + 9x + 20 = 0. The roots are [-9 ± sqrt(81 - 80)] / 2 = [-9 ± 1]/2. So, (-9 + 1)/2 = -4, (-9 -1)/2 = -5. Correct.

Therefore, the only common solution is x = -4. Therefore, the answer is x = -4.

But let me check if there's a possibility that even if the expressions inside the absolute values are not zero, their absolute values could sum to zero? But since absolute values are non-negative, the sum can only be zero if both are zero. Therefore, no.

Therefore, the conclusion is that the only real value not in the domain is x = -4. Therefore, the answer is \boxed{-4}.

Wait, but the question says "real values of x are not in the domain". So, maybe there are multiple values? But according to my analysis, only x = -4. Let me check once more.

Suppose that there is some x where |x² + 3x - 4| + |x² + 9x + 20| = 0. Since absolute values are non-negative, each term must be zero. Therefore, solving x² + 3x - 4 = 0 and x² + 9x + 20 = 0. Which gives x = -4 as the only common solution. Therefore, only x = -4.

Alternatively, maybe there's a case where one of the expressions inside the absolute value is negative? But even if they are negative, the absolute value makes them positive. Therefore, their sum can't be zero unless both are zero.

Therefore, I think my conclusion is correct. Therefore, the answer is \boxed{-4}.

**Final Answer**
The real value not in the domain of $ f(x) $ is \boxed{-4}.

To determine the real values of $ x $ that are **not in the domain** of the function

$$
f(x) = \frac{1}{|x^2 + 3x - 4| + |x^2 + 9x + 20|},
$$

we must identify the values of $ x $ for which the **denominator becomes zero**, since division by zero is undefined.

---

### Step 1: Understand the Denominator

The denominator is:

$$
|x^2 + 3x - 4| + |x^2 + 9x + 20|.
$$

Since absolute values are always **non-negative**, the **sum of two absolute values is zero** **only if both absolute values are zero**. That is:

$$
|x^2 + 3x - 4| = 0 \quad \text{and} \quad |x^2 + 9x + 20| = 0.
$$

This implies:

$$
x^2 + 3x - 4 = 0 \quad \text{and} \quad x^2 + 9x + 20 = 0.
$$

---

### Step 2: Solve the Quadratic Equations

#### First Equation: $ x^2 + 3x - 4 = 0 $

Use the quadratic formula:

$$
x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm \sqrt{25}}{2} = \frac{-3 \pm 5}{2}.
$$

So the solutions are:

$$
x = \frac{-3 + 5}{2} = 1, \quad x = \frac{-3 - 5}{2} = -4.
$$

#### Second Equation: $ x^2 + 9x + 20 = 0 $

Again, use the quadratic formula:

$$
x = \frac{-9 \pm \sqrt{9^2 - 4(1)(20)}}{2(1)} = \frac{-9 \pm \sqrt{81 - 80}}{2} = \frac{-9 \pm \sqrt{1}}{2} = \frac{-9 \pm 1}{2}.
$$

So the solutions are:

$$
x = \frac{-9 + 1}{2} = -4, \quad x = \frac{-9 - 1}{2} = -5.
$$

---

### Step 3: Find Common Solutions

We now look for values of $ x $ that satisfy **both** equations:

- First equation: $ x = 1, -4 $
- Second equation: $ x = -4, -5 $

The only common solution is:

$$
x = -4.
$$

---

### Step 4: Verify the Denominator at $ x = -4 $

Plug $ x = -4 $ into the denominator:

- First term: $ |(-4)^2 + 3(-4) - 4| = |16 - 12 - 4| = |0| = 0 $
- Second term: $ |(-4)^2 + 9(-4) + 20| = |16 - 36 + 20| = |0| = 0 $

So the denominator becomes:

$$
0 + 0 = 0,
$$

which confirms that $ x = -4 $ makes the function undefined.

---

### Step 5: Conclusion

Since the sum of two absolute values is zero **only when both are zero**, and the only value of $ x $ that satisfies both equations is $ x = -4 $, this is the **only** value not in the domain of $ f(x) $.

---

### Final Answer

$$
\boxed{-4}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the real values of x that are not in the domain of the function f(x) = 1 divided by (|x² + 3x - 4| + |x² + 9x + 20|). Hmm, the domain of a function is all the real numbers for which the function is defined. Since this function is a fraction, the denominator can't be zero. So, the values of x that make the denominator zero are not in the domain. Therefore, I need to find all real numbers x where |x² + 3x - 4| + |x² + 9x + 20| = 0. But wait, the sum of two absolute values is zero only if each absolute value is zero individually. Because absolute values are always non-negative, so their sum can only be zero if both terms are zero. So, I need to solve the system:

|x² + 3x - 4| = 0 and |x² + 9x + 20| = 0.

Which implies:

x² + 3x - 4 = 0 and x² + 9x + 20 = 0.

So, solving these two quadratic equations separately and then finding their intersection (since both must be true at the same time) will give me the values of x not in the domain. Let me start with the first equation: x² + 3x - 4 = 0.

To solve this quadratic equation, I can use the quadratic formula. The discriminant D is b² - 4ac = 9 + 16 = 25. So, the roots are (-3 ± √25)/2 = (-3 ± 5)/2. Therefore, the solutions are (-3 + 5)/2 = 2/2 = 1, and (-3 - 5)/2 = -8/2 = -4. So, the first equation has solutions x = 1 and x = -4.

Now, moving on to the second equation: x² + 9x + 20 = 0. Again, using the quadratic formula. The discriminant D is 81 - 80 = 1. So, the roots are (-9 ± √1)/2 = (-9 ± 1)/2. Therefore, the solutions are (-9 + 1)/2 = -8/2 = -4, and (-9 - 1)/2 = -10/2 = -5. So, the second equation has solutions x = -4 and x = -5.

Now, the system requires both equations to be satisfied. So, looking for x values that are in both solutions. The first equation has x = 1 and x = -4. The second equation has x = -4 and x = -5. The only common solution is x = -4. Therefore, x = -4 is the only value where both absolute value expressions are zero. Therefore, the denominator becomes zero at x = -4, so x = -4 is not in the domain of f(x).

Wait, but hold on. Let me check if there are any other possibilities. Because sometimes when dealing with absolute values, there might be cases where the expressions inside are negative, but since we have absolute values, they are always non-negative. However, the sum of two non-negative numbers is zero only if both are zero. Therefore, there are no other possibilities. Therefore, the only x that makes the denominator zero is x = -4. Therefore, the domain of f(x) is all real numbers except x = -4. Therefore, the real value not in the domain is x = -4.

But wait, let me double-check. Let me plug x = -4 into the denominator. Let's compute |(-4)^2 + 3*(-4) - 4| + |(-4)^2 + 9*(-4) + 20|.

First term: 16 - 12 - 4 = 0. Absolute value is 0. Second term: 16 - 36 + 20 = 0. Absolute value is 0. So, 0 + 0 = 0. Therefore, denominator is zero. So, x = -4 is indeed not in the domain.

But let me check if there are other x values where the denominator could be zero. Suppose, for example, that one of the absolute values is positive and the other is negative? Wait, no. Absolute values are always non-negative. So, their sum can't be negative. The only way the sum is zero is if both are zero. Therefore, there are no other solutions. Therefore, only x = -4 is excluded.

But let me think again. Maybe there's a mistake here. Let me check if there are any other points where the expressions inside the absolute values could be zero. For example, maybe if one of them is zero and the other is also zero? But that's exactly what we did. So, the only x that makes both expressions zero is x = -4.

Alternatively, maybe there's a case where one of the absolute values is zero and the other is also zero? Which is the same as above.

Alternatively, maybe there's a case where the expressions inside the absolute values are not zero, but their sum is zero? But since absolute values are non-negative, their sum can't be negative. So, the sum can be zero only if both are zero. Therefore, there are no other solutions.

Therefore, the only value not in the domain is x = -4. Therefore, the answer is x = -4.

But wait, let me check if there are any other points where the denominator could be zero. For example, suppose that one of the absolute values is positive and the other is negative? But absolute values can't be negative. So, even if one is positive and the other is positive, their sum is positive. Therefore, the denominator is always positive except when both are zero. Therefore, only x = -4 is excluded.

But let me check another value. Let me pick x = 1. Let's compute the denominator. First term: |1 + 3 - 4| = |0| = 0. Second term: |1 + 9 + 20| = |30| = 30. Therefore, denominator is 0 + 30 = 30 ≠ 0. Therefore, x = 1 is in the domain. Similarly, x = -5. Let's check. First term: |(-5)^2 + 3*(-5) - 4| = |25 - 15 - 4| = |6| = 6. Second term: |(-5)^2 + 9*(-5) + 20| = |25 - 45 + 20| = |0| = 0. Therefore, denominator is 6 + 0 = 6 ≠ 0. Therefore, x = -5 is in the domain.

Therefore, only x = -4 is excluded. Therefore, the answer is x = -4.

But wait, the question says "real values of x are not in the domain". So, maybe there are more? Let me think again. Suppose that the denominator is zero. So, |A| + |B| = 0. Which implies A = 0 and B = 0. Therefore, solving A = 0 and B = 0. Which gives x = -4. Therefore, only x = -4.

Alternatively, maybe there's a mistake in solving the quadratics? Let me check again.

First quadratic: x² + 3x - 4 = 0. The roots are x = [-3 ± sqrt(9 + 16)] / 2 = [-3 ± 5]/2. So, 2/2 = 1 and -8/2 = -4. Correct.

Second quadratic: x² + 9x + 20 = 0. The roots are [-9 ± sqrt(81 - 80)] / 2 = [-9 ± 1]/2. So, (-9 + 1)/2 = -4, (-9 -1)/2 = -5. Correct.

Therefore, the only common solution is x = -4. Therefore, the answer is x = -4.

But let me check if there's a possibility that even if the expressions inside the absolute values are not zero, their absolute values could sum to zero? But since absolute values are non-negative, the sum can only be zero if both are zero. Therefore, no.

Therefore, the conclusion is that the only real value not in the domain is x = -4. Therefore, the answer is \boxed{-4}.

Wait, but the question says "real values of x are not in the domain". So, maybe there are multiple values? But according to my analysis, only x = -4. Let me check once more.

Suppose that there is some x where |x² + 3x - 4| + |x² + 9x + 20| = 0. Since absolute values are non-negative, each term must be zero. Therefore, solving x² + 3x - 4 = 0 and x² + 9x + 20 = 0. Which gives x = -4 as the only common solution. Therefore, only x = -4.

Alternatively, maybe there's a case where one of the expressions inside the absolute value is negative? But even if they are negative, the absolute value makes them positive. Therefore, their sum can't be zero unless both are zero.

Therefore, I think my conclusion is correct. Therefore, the answer is \boxed{-4}.

**Final Answer**
The real value not in the domain of $ f(x) $ is \boxed{-4}.

To determine the real values of $ x $ that are **not in the domain** of the function

$$
f(x) = \frac{1}{|x^2 + 3x - 4| + |x^2 + 9x + 20|},
$$

we must identify the values of $ x $ for which the **denominator becomes zero**, since division by zero is undefined.

---

### Step 1: Understand the Denominator

The denominator is:

$$
|x^2 + 3x - 4| + |x^2 + 9x + 20|.
$$

Since absolute values are always **non-negative**, the **sum of two absolute values is zero** **only if both absolute values are zero**. That is:

$$
|x^2 + 3x - 4| = 0 \quad \text{and} \quad |x^2 + 9x + 20| = 0.
$$

This implies:

$$
x^2 + 3x - 4 = 0 \quad \text{and} \quad x^2 + 9x + 20 = 0.
$$

---

### Step 2: Solve the Quadratic Equations

#### First Equation: $ x^2 + 3x - 4 = 0 $

Use the quadratic formula:

$$
x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm \sqrt{25}}{2} = \frac{-3 \pm 5}{2}.
$$

So the solutions are:

$$
x = \frac{-3 + 5}{2} = 1, \quad x = \frac{-3 - 5}{2} = -4.
$$

#### Second Equation: $ x^2 + 9x + 20 = 0 $

Again, use the quadratic formula:

$$
x = \frac{-9 \pm \sqrt{9^2 - 4(1)(20)}}{2(1)} = \frac{-9 \pm \sqrt{81 - 80}}{2} = \frac{-9 \pm \sqrt{1}}{2} = \frac{-9 \pm 1}{2}.
$$

So the solutions are:

$$
x = \frac{-9 + 1}{2} = -4, \quad x = \frac{-9 - 1}{2} = -5.
$$

---

### Step 3: Find Common Solutions

We now look for values of $ x $ that satisfy **both** equations:

- First equation: $ x = 1, -4 $
- Second equation: $ x = -4, -5 $

The only common solution is:

$$
x = -4.
$$

---

### Step 4: Verify the Denominator at $ x = -4 $

Plug $ x = -4 $ into the denominator:

- First term: $ |(-4)^2 + 3(-4) - 4| = |16 - 12 - 4| = |0| = 0 $
- Second term: $ |(-4)^2 + 9(-4) + 20| = |16 - 36 + 20| = |0| = 0 $

So the denominator becomes:

$$
0 + 0 = 0,
$$

which confirms that $ x = -4 $ makes the function undefined.

---

### Step 5: Conclusion

Since the sum of two absolute values is zero **only when both are zero**, and the only value of $ x $ that satisfies both equations is $ x = -4 $, this is the **only** value not in the domain of $ f(x) $.

---

### Final Answer

$$
\boxed{-4}
$$

Solution:
$x$ is not in the domain of $f$ if the denominator is zero.  Since both absolute values are nonnegative, both must be zero for the denominator to be zero.  So

\begin{align*}
0=x^2+3x-4=(x+4)(x-1)&\Rightarrow x=-4\text{ or }x=1\\
0=x^2+9x+20=(x+4)(x+5)&\Rightarrow x=-4\text{ or }x=-5
\end{align*}

The only value of $x$ which makes both absolute values zero is $x=\boxed{-4}$.