if `y^(2)(y^(2)-6)+x^(2)-8x+24=0`, then maximum value of `sqrt(y^(4)+x^(2))` is

Explanation

To find the maximum value of \( \sqrt{y^4 + x^2} \) given the equation

\[ y^2(y^2 - 6) + x^2 - 8x + 24 = 0, \]

we can follow these steps:

### Step 1: Rewrite the equation
The given equation can be rewritten as:

\[ y^4 - 6y^2 + x^2 - 8x + 24 = 0. \]

### Step 2: Rearrange the equation
Next, we can rearrange this equation to isolate \( x^2 \):

\[ x^2 - 8x + y^4 - 6y^2 + 24 = 0. \]

### Step 3: Complete the square for \( x \)
We complete the square for the \( x \) terms:

\[ x^2 - 8x = (x - 4)^2 - 16. \]

Substituting this back into the equation gives:

\[ (x - 4)^2 - 16 + y^4 - 6y^2 + 24 = 0, \]

which simplifies to:

\[ (x - 4)^2 + y^4 - 6y^2 + 8 = 0. \]

### Step 4: Rearranging again
Rearranging this gives:

\[ (x - 4)^2 = -y^4 + 6y^2 - 8. \]

### Step 5: Analyze the right side
For \( (x - 4)^2 \) to be non-negative, we need:

\[ -y^4 + 6y^2 - 8 \geq 0. \]

This implies:

\[ y^4 - 6y^2 + 8 \leq 0. \]

### Step 6: Let \( t = y^2 \)
Let \( t = y^2 \), then we have:

\[ t^2 - 6t + 8 \leq 0. \]

### Step 7: Solve the quadratic inequality
The roots of the quadratic equation \( t^2 - 6t + 8 = 0 \) can be found using the quadratic formula:

\[ t = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2}. \]

This gives us the roots:

\[ t = 4 \quad \text{and} \quad t = 2. \]

### Step 8: Determine the intervals
The quadratic opens upwards (since the coefficient of \( t^2 \) is positive), so the inequality \( t^2 - 6t + 8 \leq 0 \) holds for:

\[ 2 \leq t \leq 4. \]

### Step 9: Find \( y \)
Since \( t = y^2 \), we have:

\[ \sqrt{2} \leq |y| \leq 2. \]

### Step 10: Find \( x \)
Now substituting back to find \( x \):

From \( (x - 4)^2 = -y^4 + 6y^2 - 8 \), we need to evaluate \( -y^4 + 6y^2 - 8 \) at the endpoints \( y^2 = 2 \) and \( y^2 = 4 \).

1. For \( y^2 = 2 \):
\[ -2^2 + 6 \cdot 2 - 8 = -4 + 12 - 8 = 0 \implies (x - 4)^2 = 0 \implies x = 4. \]

2. For \( y^2 = 4 \):
\[ -4^2 + 6 \cdot 4 - 8 = -16 + 24 - 8 = 0 \implies (x - 4)^2 = 0 \implies x = 4. \]

### Step 11: Calculate \( \sqrt{y^4 + x^2} \)
Now we calculate \( \sqrt{y^4 + x^2} \) at these points:

- At \( y^2 = 2 \) and \( x = 4 \):
\[ \sqrt{(2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}. \]

- At \( y^2 = 4 \) and \( x = 4 \):
\[ \sqrt{(4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}. \]

### Step 12: Find the maximum value
Comparing \( 2\sqrt{5} \) and \( 4\sqrt{2} \):

- \( 2\sqrt{5} \approx 4.472 \)
- \( 4\sqrt{2} \approx 5.656 \)

Thus, the maximum value of \( \sqrt{y^4 + x^2} \) is:

\[ \boxed{4\sqrt{2}}. \]