If the solution of the equation `|(x^4-9) -(x^2 + 3)| = |x^4 - 9| - |x^2 + 3|` is `(-oo,p]uu[q,oo)` then

Options

A. `p+q=0`
B. `|p|+|q|=4`
C. `|q-p|=4`
D. `|pq|=4`

Answer

To solve the equation \( |(x^4 - 9) - (x^2 + 3)| = |x^4 - 9| - |x^2 + 3| \), we will break it down step by step.

### Step 1: Rewrite the equation
The equation can be rewritten as:
\[ |(x^4 - 9) - (x^2 + 3)| = |x^4 - 9| - |x^2 + 3| \]
Let \( a = x^4 - 9 \) and \( b = x^2 + 3 \). Then the equation becomes:
\[ |a - b| = |a| - |b| \]

### Step 2: Analyze the modulus condition
The property of absolute values states that \( |a - b| = |a| - |b| \) holds true if either:
1. \( a \geq b \) and both \( a \) and \( b \) have the same sign.
2. \( a < b \) and both \( a \) and \( b \) have the same sign.

### Step 3: Consider the first case \( a \geq b \)
From \( a \geq b \):
\[ x^4 - 9 \geq x^2 + 3 \]
Rearranging gives:
\[ x^4 - x^2 - 12 \geq 0 \]
Factoring:
\[ (x^2 - 4)(x^2 + 3) \geq 0 \]
Since \( x^2 + 3 > 0 \) for all \( x \), we only need to consider:
\[ x^2 - 4 \geq 0 \implies (x - 2)(x + 2) \geq 0 \]
The critical points are \( x = -2 \) and \( x = 2 \). The intervals to test are:
- \( (-\infty, -2) \)
- \( (-2, 2) \)
- \( (2, \infty) \)

Testing these intervals:
- For \( x < -2 \): both factors are positive, so the product is positive.
- For \( -2 < x < 2 \): one factor is negative, so the product is negative.
- For \( x > 2 \): both factors are positive, so the product is positive.

Thus, the solution for this case is:
\[ (-\infty, -2] \cup [2, \infty) \]

### Step 4: Consider the second case \( a < b \)
From \( a < b \):
\[ x^4 - 9 < x^2 + 3 \]
Rearranging gives:
\[ x^4 - x^2 - 12 < 0 \]
Factoring gives the same expression:
\[ (x^2 - 4)(x^2 + 3) < 0 \]
Again, since \( x^2 + 3 > 0 \), we only consider:
\[ x^2 - 4 < 0 \implies -2 < x < 2 \]

### Step 5: Combine results
From the first case, we have:
\[ (-\infty, -2] \cup [2, \infty) \]
From the second case, we have:
\[ (-2, 2) \]
Combining these results, we find the solution set:
\[ (-\infty, -2] \cup [2, \infty) \]

### Final Result
The solution of the equation is:
\[ (-\infty, p] \cup [q, \infty) \quad \text{where } p = -2 \text{ and } q = 2 \]