Question

Show work when rationalizing the denomiator:

Answer 1

To rationalize the denominator, multiply both numerator and denominator by its conjugate, which is \(2+\sqrt{3}\):

\[\frac{\sqrt{6}}{2-\sqrt{3}} = \frac{(2+\sqrt{3})}{(2+\sqrt{3})}\frac{\sqrt{6}}{(2-\sqrt{3})}\]

Answer 2

That will give you a denominator which is a difference of squares: \((2^2-(\sqrt{3})^2)\) and you can simplify from there.

Answer 3

\[\frac{ \sqrt6 }{ 2-\sqrt3 } \times \frac{ 2+\sqrt3 }{ 2+\sqrt3 }\]
\[\frac{ \sqrt6(2+\sqrt3) }{ (2-\sqrt3)(2+\sqrt3) }\]
using the fact that
\[(a-b)(a+b)=a^2-b^2\]
\[\frac{ 2\sqrt6+\sqrt{18} }{ (2)^2-(\sqrt3)^2 }\]
\[\frac{ 2\sqrt6 + \sqrt{9 \times 2}}{ 4-3 }\]
\[\frac{ 2\sqrt6+3\sqrt2 }{ 1 } ~=~2\sqrt6+3\sqrt2\]

Answer 4

@whpalmer4 Would simplifying (2^2−(3√)^2) give me (4-2.44)? And @kausarsalley is that the complete rationalization?

Answer 5

No, \(2^2-(\sqrt{3})^2 = 2*2 - \sqrt{3}*\sqrt{3} = 4 - 3 = 1\)

Answer 6

You'll use this multiplying by the conjugate technique a lot, so get used to it :-)

Trig identities, complex numbers, rationalizing radicals, etc.

Answer 7

Yeah, @kausarsalley's solution is complete, though it could also be written as \(\sqrt{2}(3+2\sqrt{3})\)