The number of real roots of the equation `(x+3)^(4)+(x+5)^(4)=16` is

Options

A. 0
B. 2
C. 4
D. none

Answer

To solve the equation \((x+3)^4 + (x+5)^4 = 16\) and determine the number of real roots, we can follow these steps:

### Step 1: Substitute Variables
Let \(a = x + 4\). Then we can rewrite \(x + 3\) and \(x + 5\) in terms of \(a\):
- \(x + 3 = a - 1\)
- \(x + 5 = a + 1\)

Thus, the equation becomes:
\[ (a - 1)^4 + (a + 1)^4 = 16 \]

### Step 2: Expand the Terms
Now we expand \((a - 1)^4\) and \((a + 1)^4\):
\[ (a - 1)^4 = a^4 - 4a^3 + 6a^2 - 4a + 1 \]
\[ (a + 1)^4 = a^4 + 4a^3 + 6a^2 + 4a + 1 \]

### Step 3: Combine the Expanded Terms
Adding these two expansions together:
\[ (a - 1)^4 + (a + 1)^4 = (a^4 - 4a^3 + 6a^2 - 4a + 1) + (a^4 + 4a^3 + 6a^2 + 4a + 1) \]
This simplifies to:
\[ 2a^4 + 12a^2 + 2 = 16 \]

### Step 4: Rearrange the Equation
Now we rearrange the equation:
\[ 2a^4 + 12a^2 + 2 - 16 = 0 \]
\[ 2a^4 + 12a^2 - 14 = 0 \]
Dividing the entire equation by 2:
\[ a^4 + 6a^2 - 7 = 0 \]

### Step 5: Substitute \(b = a^2\)
Let \(b = a^2\). Then the equation becomes:
\[ b^2 + 6b - 7 = 0 \]

### Step 6: Solve the Quadratic Equation
Now we can solve this quadratic equation using the quadratic formula:
\[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
where \(A = 1\), \(B = 6\), and \(C = -7\):
\[ b = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} \]
\[ b = \frac{-6 \pm \sqrt{36 + 28}}{2} \]
\[ b = \frac{-6 \pm \sqrt{64}}{2} \]
\[ b = \frac{-6 \pm 8}{2} \]

### Step 7: Calculate the Values of \(b\)
Calculating the two possible values for \(b\):
1. \(b = \frac{2}{2} = 1\)
2. \(b = \frac{-14}{2} = -7\) (not valid since \(b = a^2\) cannot be negative)

### Step 8: Find Values of \(a\)
Since \(b = 1\):
\[ a^2 = 1 \implies a = 1 \text{ or } a = -1 \]

### Step 9: Find Values of \(x\)
Recall that \(a = x + 4\):
1. If \(a = 1\), then \(x + 4 = 1 \implies x = -3\)
2. If \(a = -1\), then \(x + 4 = -1 \implies x = -5\)

### Conclusion
Thus, the equation \((x+3)^4 + (x+5)^4 = 16\) has **two real roots**: \(x = -3\) and \(x = -5\).

### Final Answer
The number of real roots of the equation is **2**.

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