# How to rationalize this denominator?

How would one rationalize $\frac6{(1-\sqrt3)}$?

My answer is $\frac{6+6\sqrt3}{-2}$ ? Is this correct? Not sure its in the simplified form..? Simple pre-calc but apparently I am stuck. Thank you.

• You can note that $6/(-2)=-3$ – egreg Jan 25 '18 at 22:35

• $$\frac{6+6\sqrt3}{-2}$$ is correct but observe that you can still cancel $2$ from the numerator and the denominator.
• The trick to solve the question is to multiply $1+\sqrt3$ (i.e. the conjugate to both the top and bottom) and use the formula of $(a-b)(a+b)=a^2-b^2$.
• hmm... not really. there shouldn't be any $6$ in the numerator in the end. also you can multiply $-1$ to both the numerator and denominator. – Siong Thye Goh Jan 25 '18 at 22:39
$\frac6{(1-\sqrt3)}*\frac{1+\sqrt3}{1+\sqrt3} = \frac{6+6\sqrt3}{1-3} = \frac{-6-6\sqrt3}{2} = -3-3\sqrt{3}$