Rationalize the denominator. $\frac { \sqrt { 3 } + \sqrt { 2 } } { \sqrt { 3 } - \sqrt { 2 } }$
A) $- 5 - 2 \sqrt { 6 }$
B) $5 + 2 \sqrt { 6 }$
C) $5 \sqrt { 6 } + 1$
D) $- 5 + 2 \sqrt { 6 }$
E) $5 - 2 \sqrt { 6 }$

Question

Quizplus · Accepted Answer

To rationalize the denominator of $\frac { \sqrt { 3 } + \sqrt { 2 } } { \sqrt { 3 } - \sqrt { 2 } }$, multiply both the numerator and the denominator by the conjugate of the denominator, which is $\sqrt { 3 } + \sqrt { 2 }$. This results in $\frac { (\sqrt { 3 } + \sqrt { 2 })^2 } { (\sqrt { 3 })^2 - (\sqrt { 2 })^2 }$, simplifying to $\frac { 3 + 2\sqrt { 6 } + 2 } { 3 - 2 }$ = $5 + 2\sqrt { 6 }$.

Rationalize the Denominator 3+23−2\frac { \sqrt { 3 } + \sqrt { 2 } } { \sqrt { 3 } - \sqrt { 2 } }3​−2​3​+2​​

Rationalize the Denominator $\frac { \sqrt { 3 } + \sqrt { 2 } } { \sqrt { 3 } - \sqrt { 2 } }$