Identifying Quadratic Equation Equivalence

The goal is to find which equation, upon simplification, reduces to a quadratic equation, which is an equation of the form $ax^2 + bx + c = 0$ with $a \neq 0$. We will analyze each option.

Equation Analysis

Option 1: $x^3 - 3x^2 + 3x - 1 = (x - 1)^3$

Expanding the right side gives $(x - 1)^3 = x^3 - 3x^2 + 3x - 1$. The equation becomes $x^3 - 3x^2 + 3x - 1 = x^3 - 3x^2 + 3x - 1$, which simplifies to $0 = 0$. This is an identity, true for all $x$, not a quadratic equation.

Option 2: $\frac{x}{x - 1} + \frac{2x}{x - 1} = 3$

Combine the terms on the left:

• $ \frac{x + 2x}{x - 1} = 3 $
• $ \frac{3x}{x - 1} = 3 $

Assuming $x \neq 1$, multiply by $(x-1)$: $3x = 3(x-1) \implies 3x = 3x - 3 \implies 0 = -3$. This is a contradiction, indicating no solution.

Option 3: $(x^2 + 1)^2 - x^4 = 2x + 1$

Expand the squared term:

• $ (x^4 + 2x^2 + 1) - x^4 = 2x + 1 $

Simplify:

• $ 2x^2 + 1 = 2x + 1 $
• $ 2x^2 = 2x $
• $ 2x^2 - 2x = 0 $

This equation is in the form $ax^2 + bx + c = 0$ where $a=2$, $b=-2$, and $c=0$. Thus, it is a quadratic equation.

Option 4: $\frac{(x^2 - 1)}{(x + 1)} = x - 2$

Factor the numerator $x^2 - 1 = (x-1)(x+1)$:

• $ \frac{(x - 1)(x + 1)}{(x + 1)} = x - 2 $

Assuming $x \neq -1$, cancel $(x+1)$: $x - 1 = x - 2 \implies -1 = -2$. This is a contradiction, indicating no solution.

Conclusion

Only Option 3 simplifies to a quadratic equation ($2x^2 - 2x = 0$).