The equation `(6-x)^(4)+(8-x)^(4)=16` has

Options

A. sum of roots 28
B. product of roots 2688
C. two real roots
D. two imaginary roots

Answer

To solve the equation \((6-x)^{4} + (8-x)^{4} = 16\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ (6-x)^{4} + (8-x)^{4} = 16 \]
We can express \(6\) as \(7-1\) and \(8\) as \(7+1\):
\[ (7-1-x)^{4} + (7+1-x)^{4} = 16 \]
This simplifies to:
\[ (7-x-1)^{4} + (7-x+1)^{4} = 16 \]
Let \(p = 7 - x\). Then, we can rewrite the equation as:
\[ (p-1)^{4} + (p+1)^{4} = 16 \]

### Step 2: Expand the equation
Now we expand \((p-1)^{4}\) and \((p+1)^{4}\):
\[ (p-1)^{4} = p^{4} - 4p^{3} + 6p^{2} - 4p + 1 \]
\[ (p+1)^{4} = p^{4} + 4p^{3} + 6p^{2} + 4p + 1 \]
Adding these two expansions together:
\[ (p-1)^{4} + (p+1)^{4} = (p^{4} - 4p^{3} + 6p^{2} - 4p + 1) + (p^{4} + 4p^{3} + 6p^{2} + 4p + 1) \]
This simplifies to:
\[ 2p^{4} + 12p^{2} + 2 = 16 \]

### Step 3: Simplify the equation
Now, we can simplify the equation:
\[ 2p^{4} + 12p^{2} + 2 - 16 = 0 \]
\[ 2p^{4} + 12p^{2} - 14 = 0 \]
Dividing the entire equation by 2:
\[ p^{4} + 6p^{2} - 7 = 0 \]

### Step 4: Substitute \(y\) for \(p^{2}\)
Let \(y = p^{2}\). Then, we have:
\[ y^{2} + 6y - 7 = 0 \]

### Step 5: Solve the quadratic equation
Now we can use the quadratic formula to solve for \(y\):
\[ y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 6\), and \(c = -7\):
\[ y = \frac{-6 \pm \sqrt{6^{2} - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} \]
\[ y = \frac{-6 \pm \sqrt{36 + 28}}{2} \]
\[ y = \frac{-6 \pm \sqrt{64}}{2} \]
\[ y = \frac{-6 \pm 8}{2} \]
Calculating the two possible values for \(y\):
1. \(y = \frac{2}{2} = 1\)
2. \(y = \frac{-14}{2} = -7\) (not valid since \(y = p^{2}\) must be non-negative)

So, we have:
\[ p^{2} = 1 \implies p = 1 \text{ or } p = -1 \]

### Step 6: Solve for \(x\)
Recall that \(p = 7 - x\):
1. If \(p = 1\):
\[ 7 - x = 1 \implies x = 6 \]
2. If \(p = -1\):
\[ 7 - x = -1 \implies x = 8 \]

### Conclusion
The equation \((6-x)^{4} + (8-x)^{4} = 16\) has two solutions:
\[ x = 6 \quad \text{and} \quad x = 8 \]