Question 118008
1) Perform the indicated operations:

{{{sqrt(50)+2sqrt(32)-sqrt(8)}}}
The key here is to factor the integers under the radical (the radicands) and see if you have any perfect squares that can be taken outside the radical by taking the square root.
{{{sqrt(50) = sqrt(2*25)}}} 25 is a perfect square, {{{25 = 5^2}}} and {{{sqrt(5^2) = 5}}}, so...
{{{sqrt(50) = sqrt(2*25)}}}={{{sqrt(2)*sqrt(25)}}}={{{sqrt(2)*5}}} or {{{5*sqrt(2)}}}
Similarly for:
{{{2*sqrt(32) = 2*sqrt(2*16)}}} = {{{2*sqrt(2)*sqrt(16) = 2*sqrt(2)*4}}}or {{{2*4*sqrt(2) = 8*sqrt(2)}}}
and for:
{{{sqrt(8) = sqrt(2*4)}}}={{{sqrt(2)*sqrt(4) = sqrt(2)*2}}} or {{{2*sqrt(2)}}}
OK, now let's put all of this together:
{{{sqrt(50)+2sqrt(32)-sqrt(8) = 5*sqrt(2)+8*sqrt(2)-2*sqrt(2)}}} Now we can combine all of the {{{sqrt(2)}}}'s
{{{5*sqrt(2)+8*sqrt(2)-2*sqrt(2) = (5+8-2)*sqrt(2)}}}={{{11sqrt(2)}}}

2) Rationalize the denominator:
{{{2/(sqrt(6)-sqrt(5))}}}
To rationalize the denominator, you will multiply both the numerator and the denominator by the "conjugate" of the denominator {{{sqrt(6)-sqrt(5)}}} and this is just {{{sqrt(6)+sqrt(5)}}}
The conjugate of (a+b) is (a-b) and the conjugate of (a-b) is (a+b), see the idea?
{{{(2/(sqrt(6)-sqrt(5)))*((sqrt(6)+sqrt(5))/(sqrt(6)+sqrt(5)))=2*(sqrt(6)+sqrt(5))/((sqrt(6)-sqrt(5))*(sqrt(6)+sqrt(5)))}}}
Let's look at the denominator which is in the form of {{{(a-b)(a+b) = a^2-b^2}}}
So after multiplying, as indicated, the denominator will look like:
{{{(sqrt(6))^2 - (sqrt(5))^2 = 6-5}}}={{{1}}}, so all we have left is the numerator:
{{{2*(sqrt(6)+sqrt(5))}}}