[tex]\frac{6}{\sqrt{3} + 2}[/tex]
When we're dealing with surdic equations on the denominator, we want to rationalise this. When we have two terms, one being the surd, in the denominator, we multiply it by the conjugate, such that the denominator becomes a difference of two squares.
[tex]\frac{6}{\sqrt{3} + 2} \cdot \frac{\sqrt{3} - 2}{\sqrt{3} - 2}[/tex]
This is still equivalent to the original equation, because we are essentially multiplying by a factor of 1.
[tex]\frac{6}{\sqrt{3} + 2} \cdot \frac{\sqrt{3} - 2}{\sqrt{3} - 2}[/tex]
[tex]= \frac{6\sqrt{3} - 12}{3 - 4}[/tex]
[tex]= -6\sqrt{3} + 12[/tex]
[tex]= 6(2 - \sqrt{3})[/tex] is our simplified form.
