Question

Simplify by rationalising the denominator: $\frac{6-4\sqrt{3}}{6+4\sqrt{3}}$.

• A. $7-4\sqrt{3}$
• B. $4\sqrt{3}-7$
• C. $7-16\sqrt{3}$
• D. $84-4\sqrt{3}$

Answer 1

Answer: (B)

$4\sqrt{3}-7$

$\left(\frac{6-4\sqrt{3}}{6+4\sqrt{3}}\right)\cdot \left(\frac{6-4\sqrt{3}}{6-4\sqrt{3}}\right)=\left(\frac{(6-4\sqrt{3}{)}^2}{36-48}\right)=\left(\frac{36+48-48\sqrt{3}}{-12}\right)=\left(\frac{84-48\sqrt{3}}{-12}\right)=-7+4\sqrt{3}=4\sqrt{3}-7$