Question

How can you write the expression with a rationalized denominator? square root of three minus six over square root of three plus six

Answer 1

Answer:

$- (\frac{15 - 4\sqrt{3} }{11} )$

Step-by-step explanation:

$\frac{\sqrt{3}-6}{\sqrt{3} +6} = \frac{(\sqrt{3}-6)(\sqrt{3}-6)}{(\sqrt{3} +6)(\sqrt{3} -6)}$

$= \frac{(\sqrt{3})^2 - (2\times6\times\sqrt{3}) +6^{2} }{(\sqrt{3})^2 - 6^{2}} = \frac{9 - 12\sqrt{3} + 36}{3 - 36}$

$\frac{45 -12\sqrt{3} }{-33} =- (\frac{15 - 4\sqrt{3} }{11} )$