If the real numbers x, y, z are such that `x^2 + 4y^2 + 16z^2 = 48` and `xy + 4yz + 2zx = 24`. what is the value of `x^(2) +y^(2)+ z^(2)`=?

Answer

To solve the problem, we need to find the value of \( x^2 + y^2 + z^2 \) given the equations:

1. \( x^2 + 4y^2 + 16z^2 = 48 \) (Equation 1)
2. \( xy + 4yz + 2zx = 24 \) (Equation 2)

### Step 1: Rewrite the equations
We have the two equations as given above. We will use them to eliminate some variables.

### Step 2: Multiply Equation 2 by 2
Multiply the second equation by 2 to help eliminate terms when we subtract:
\[ 2(xy + 4yz + 2zx) = 2 \cdot 24 \]
This simplifies to:
\[ 2xy + 8yz + 4zx = 48 \quad \text{(Equation 3)} \]

### Step 3: Subtract Equation 3 from Equation 1
Now, we will subtract Equation 3 from Equation 1:
\[ (x^2 + 4y^2 + 16z^2) - (2xy + 8yz + 4zx) = 48 - 48 \]
This simplifies to:
\[ x^2 + 4y^2 + 16z^2 - 2xy - 8yz - 4zx = 0 \]

### Step 4: Rearranging the terms
Rearranging the equation, we can group the terms:
\[ x^2 - 2xy + 4y^2 - 8yz + 16z^2 - 4zx = 0 \]

### Step 5: Completing the square
Now we will complete the square for each group of terms:
1. For \( x^2 - 2xy \), we can write it as \( (x - 2y)^2 \).
2. For \( 4y^2 - 8yz + 16z^2 \), we can write it as \( (2y - 4z)^2 \).
3. For \( 16z^2 - 4zx \), we can write it as \( (4z - x)^2 \).

Thus, we have:
\[ (x - 2y)^2 + (2y - 4z)^2 + (4z - x)^2 = 0 \]

### Step 6: Setting each square to zero
Since the sum of squares equals zero, each individual square must also equal zero:
\[ x - 2y = 0 \quad (1) \]
\[ 2y - 4z = 0 \quad (2) \]
\[ 4z - x = 0 \quad (3) \]

### Step 7: Solving the equations
From equation (1):
\[ x = 2y \]
From equation (2):
\[ y = 2z \]
From equation (3):
\[ z = \frac{x}{4} \]

### Step 8: Substituting back to find values
Let \( z = k \). Then:
\[ y = 2k \quad \text{and} \quad x = 2y = 4k \]

### Step 9: Substitute into Equation 1
Now substitute \( x, y, z \) into Equation 1:
\[ (4k)^2 + 4(2k)^2 + 16(k)^2 = 48 \]
This simplifies to:
\[ 16k^2 + 16k^2 + 16k^2 = 48 \]
\[ 48k^2 = 48 \]
\[ k^2 = 1 \quad \Rightarrow \quad k = 1 \quad \text{or} \quad k = -1 \]

### Step 10: Finding \( x, y, z \)
If \( k = 1 \):
\[ z = 1, \quad y = 2, \quad x = 4 \]
If \( k = -1 \):
\[ z = -1, \quad y = -2, \quad x = -4 \]

### Step 11: Calculate \( x^2 + y^2 + z^2 \)
Now we calculate:
\[ x^2 + y^2 + z^2 = 4^2 + 2^2 + 1^2 = 16 + 4 + 1 = 21 \]

Thus, the value of \( x^2 + y^2 + z^2 \) is:
\[ \boxed{21} \]