This problem involves finding the quadratic equation with rational coefficients when one root is given as $x_1 = 2 - \sqrt{3}$.

Step 1: Understanding the Roots of Quadratic Equations

If a quadratic equation has rational coefficients and one root is $x_1 = 2 - \sqrt{3}$, the other root, $x_2$, must be its conjugate, $x_2 = 2 + \sqrt{3}$, to ensure the coefficients remain rational.

Step 2: Form the Quadratic Equation

The quadratic equation can be expressed as: $(x - x_1)(x - x_2) = 0$ Substitute $x_1 = 2 - \sqrt{3}$ and $x_2 = 2 + \sqrt{3}$: $(x - (2 - \sqrt{3}))(x - (2 + \sqrt{3})) = 0$

Step 3: Simplify the Expression

Expand the terms: $(x - 2 + \sqrt{3})(x - 2 - \sqrt{3}) = 0$ This is a difference of squares: $(x - 2)^2 - (\sqrt{3})^2 = 0$ Simplify further: $(x - 2)^2 - 3 = 0$ Expand $(x - 2)^2$: $x^2 - 4x + 4 - 3 = 0$ Simplify: $x^2 - 4x + 1 = 0$

Step 4: Match the Correct Option

From the given options, the correct answer is: C) $x^2 - 4x + 1 = 0$

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Related Questions:

1. How do you determine the conjugate of an irrational root?
2. Why must coefficients of quadratic equations be rational when the roots include a surd?
3. What are some common methods to solve quadratic equations?
4. How do you verify that a quadratic equation satisfies both roots?
5. What happens to the symmetry of roots if a different quadratic coefficient is selected?

Tip:

Always remember that for a quadratic equation with rational coefficients, irrational roots must occur in conjugate pairs.