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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.

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  • $\begingroup$ You can note that $6/(-2)=-3$ $\endgroup$ – egreg Jan 25 '18 at 22:35
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  • Do not omit braces

  • $$\frac{6+6\sqrt3}{-2}$$ 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 $(a-b)(a+b)=a^2-b^2$.

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  • $\begingroup$ So what would be the full simplified answer? 3+6sqrt(3)/-1? $\endgroup$ – user522534 Jan 25 '18 at 22:37
  • $\begingroup$ 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. $\endgroup$ – Siong Thye Goh Jan 25 '18 at 22:39
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$\frac6{(1-\sqrt3)}*\frac{1+\sqrt3}{1+\sqrt3} = \frac{6+6\sqrt3}{1-3} = \frac{-6-6\sqrt3}{2} = -3-3\sqrt{3}$

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