Solving for x in Quadratic Equations and Evaluating Complex Expressions

The problem presented provides a compelling insight into the nature of quadratic equations and their solutions, as well as the elegant process of evaluating complex polynomial expressions. By tackling the equation ( x^2 - 2x 2 0 ), we delve into the intricacies of algebraic manipulation and the application of the quadratic formula.

Step-by-Step Solution of the Quadratic Equation

Given the equation ( x^2 - 2x 2 0 ), our first step is to rearrange the equation for clarity:

( x^2 - 2x - 2 0 )

To solve for x, we use the quadratic formula:

( x frac{-b pm sqrt{b^2 - 4ac}}{2a} )

Here, a 1, b -2, and c -2. Plugging these values into the formula, we get:

( x frac{2 pm sqrt{4 - 8}}{2} )

Further simplifying:

( x frac{2 pm sqrt{-4}}{2} )

Given that the square root of a negative number results in an imaginary number, we have:

( x frac{2 pm 2i}{2} ), which simplifies to:

( x 1 pm i )

This shows that the solutions for x are imaginary numbers, specifically 1 i and 1 - i.

Evaluating the Expression ( x^4 - x^3 x^2 - 2 )

Next, we tackle the evaluation of the expression ( x^4 - x^3 x^2 - 2 ) using the fact that ( x^2 2x - 2 ):

Step 1: Calculate ( x^3 )

( x^3 x x^2 x (2x - 2) ),

Substituting ( x^2 2x - 2 ), we get:

( x^3 2x^2 - 2x 2 (2x - 2) - 2x 4x - 4 - 2x 2x - 4 )

Step 2: Calculate ( x^4 )

( x^4 x x^3 x (2x - 4) ).

Substituting ( x^2 2x - 2 ), we get:

( x^4 2x^2 - 4x 2 (2x - 2) - 4x 4x - 4 - 4x -4 )

Step 3: Substituting into the Expression

Now, substituting ( x^4 -4 ), ( x^3 2x - 4 ), and ( x^2 2x - 2 ) into the original expression:

( x^4 - x^3 x^2 - 2 -4 - (2x - 4) (2x - 2) - 2 )

Further simplifying:

( -4 - (4x^2 - 4x - 4x 8) - 2 )

( -4 - (4x^2 - 8x 8) - 2 )

( -4 - (4 (2x - 2) - 8x 8) - 2 )

( -4 - (8x - 8 - 8x 8) - 2 )

( -4 - 0 - 2 -6 )

However, simplifying it further, we find:

( -4 - -4 - 2 0 )

Therefore, the final result is:

( -4 - (2x - 4)(2x - 2) - 2 0 )

Final Answer: ( boxed{0} )

Conclusion

This comprehensive solution not only shows how to solve a quadratic equation and evaluate a complex polynomial expression but also highlights the importance of algebraic manipulation and the application of fundamental algebraic principles.