Integral part of the product of non-real roots of equation `x^(4)-4x^(3)+6x^(2)-4x=69` is _______.

Explanation

To find the integral part of the product of the non-real roots of the equation \(x^4 - 4x^3 + 6x^2 - 4x = 69\), we can follow these steps:

### Step 1: Rearranging the Equation
First, we rearrange the given equation:
\[ x^4 - 4x^3 + 6x^2 - 4x - 69 = 0 \]

### Step 2: Completing the Square
Next, we can complete the square for the polynomial on the left side. Notice that the first three terms can be grouped:
\[ x^4 - 4x^3 + 6x^2 = (x^2 - 2x)^2 + 2 \]
Thus, we can rewrite the equation as:
\[ (x^2 - 2x)^2 + 2 - 4x - 69 = 0 \]
This simplifies to:
\[ (x^2 - 2x + 1)^2 - 70 = 0 \]
or
\[ (x - 1)^4 = 70 \]

### Step 3: Taking the Square Root
Taking the square root of both sides gives:
\[ x - 1 = \pm \sqrt[4]{70} \]
Thus, we have:
\[ x = 1 \pm \sqrt[4]{70} \]

### Step 4: Finding the Non-Real Roots
The roots of the equation can be expressed as:
1. \(x = 1 + \sqrt[4]{70}\)
2. \(x = 1 - \sqrt[4]{70}\)

To find the non-real roots, we need to consider the quadratic form derived from the equation:
\[ (x - 1)^2 = -\sqrt{70} \]
This leads to:
\[ x^2 - 2x + 1 + \sqrt{70} = 0 \]

### Step 5: Finding the Product of Non-Real Roots
The product of the roots of the quadratic equation \(ax^2 + bx + c = 0\) is given by \(\frac{c}{a}\). Here, we have:
\[ c = 1 + \sqrt{70}, \quad a = 1 \]
Thus, the product of the non-real roots is:
\[ 1 + \sqrt{70} \]

### Step 6: Calculating the Integral Part
Now, we need to calculate \(1 + \sqrt{70}\):
\[ \sqrt{70} \approx 8.37 \quad (\text{since } 8^2 = 64 \text{ and } 9^2 = 81) \]
So,
\[ 1 + \sqrt{70} \approx 1 + 8.37 = 9.37 \]
The integral part of this value is:
\[ \lfloor 9.37 \rfloor = 9 \]

### Final Answer
Thus, the integral part of the product of the non-real roots of the equation is:
\[ \boxed{9} \]