The number of real solutions of the equation `root4(97-x) + root4(x) =5`
(1) 0 (2) 1
(3) 2 (4) 4

Answer

To find the number of real solutions for the equation \( \sqrt[4]{97 - x} + \sqrt[4]{x} = 5 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sqrt[4]{97 - x} + \sqrt[4]{x} = 5 \]
Let \( y_1 = \sqrt[4]{97 - x} \) and \( y_2 = \sqrt[4]{x} \). Thus, we can rewrite the equation as:
\[ y_1 + y_2 = 5 \]

### Step 2: Express \( y_1 \) in terms of \( y_2 \)
From the equation \( y_1 + y_2 = 5 \), we can express \( y_1 \) as:
\[ y_1 = 5 - y_2 \]

### Step 3: Substitute back to find \( x \)
Now, substituting back for \( y_1 \) and \( y_2 \):
\[ \sqrt[4]{97 - x} = 5 - \sqrt[4]{x} \]
Raising both sides to the power of 4 gives:
\[ 97 - x = (5 - \sqrt[4]{x})^4 \]

### Step 4: Expand the right-hand side
Next, we need to expand \( (5 - \sqrt[4]{x})^4 \). Using the binomial theorem:
\[ (5 - \sqrt[4]{x})^4 = 5^4 - 4 \cdot 5^3 \cdot \sqrt[4]{x} + 6 \cdot 5^2 \cdot (\sqrt[4]{x})^2 - 4 \cdot 5 \cdot (\sqrt[4]{x})^3 + (\sqrt[4]{x})^4 \]
Calculating \( 5^4 = 625 \), \( 4 \cdot 5^3 = 500 \), \( 6 \cdot 5^2 = 150 \), and \( 4 \cdot 5 = 20 \), we get:
\[ (5 - \sqrt[4]{x})^4 = 625 - 500\sqrt[4]{x} + 150\sqrt{x} - 20x^{3/4} + x \]

### Step 5: Set up the equation
Now we can set up the equation:
\[ 97 - x = 625 - 500\sqrt[4]{x} + 150\sqrt{x} - 20x^{3/4} + x \]
Rearranging gives:
\[ 0 = 625 - 97 - 500\sqrt[4]{x} + 150\sqrt{x} - 20x^{3/4} + 2x \]
This simplifies to:
\[ 0 = 528 - 500\sqrt[4]{x} + 150\sqrt{x} - 20x^{3/4} + 2x \]

### Step 6: Analyze the function
Let \( t = \sqrt[4]{x} \). Then \( x = t^4 \), and we substitute:
\[ 0 = 528 - 500t + 150t^2 - 20t^3 + 2t^4 \]
This is a polynomial equation in \( t \).

### Step 7: Determine the number of real solutions
To find the number of real solutions, we can analyze the polynomial \( 2t^4 - 20t^3 + 150t - 500 + 528 = 0 \). We can use the discriminant or graphing techniques to find the number of real roots.

### Step 8: Conclusion
After analyzing the polynomial, we find that it has two real solutions for \( t \), which correspond to two values of \( x \).

Thus, the number of real solutions to the original equation is:
\[ \boxed{2} \]