Rationalize the denominator and simplify the radical expression.
$\frac{2-\sqrt{3}}{2+\sqrt{3}}$

• A. $7-4\sqrt3$
• B. $\frac{1}{7+4\sqrt3}$
• C. $7+4\sqrt3$
• D. $\frac{1}{7-4\sqrt3}$

Answer: (A)

Verified step by step guidance

1. Identify the given expression: \(\frac{2-\sqrt{3}}{2+\sqrt{3}}\). The goal is to rationalize the denominator, which means eliminating the square root from the denominator.
2. Multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \$2+\sqrt{3}\( is \)2-\sqrt{3}\(. So multiply by \)\frac{2-\sqrt{3}}{2-\sqrt{3}}$:
3. \[\frac{2-\sqrt{3}}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}\]
4. Use the difference of squares formula for the denominator: \((a+b)(a-b) = a^2 - b^2\). Here, \(a=2\) and \(b=\sqrt{3}\), so the denominator becomes \$2^2 - (\sqrt{3})^2$.
5. Expand the numerator by distributing: \((2-\sqrt{3})(2-\sqrt{3})\). Then simplify both numerator and denominator by performing the arithmetic and combining like terms.