Question

Rationalize the denominator of sqrt -49 over (7 - 2i) - (4 + 9i)

Answer 1

$\sqrt{ \frac{-49}{(7-2i)-(4+9i) } }$

This one is quite the deal, but we can begin by distributing the negative on the denominator and getting rid of the parenthesis:

$\frac{ \sqrt{-49}}{7-2i-4-9i}$

See how the denominator now is more a simplification of like terms, with this I mean that you operate the numbers with an "i" together and the ones that do not have an "i" together as well. Namely, the 7 and the -4, the -2i with the -9i.
Therefore having the result: 

$\frac{ \sqrt{-49} }{3-11i}$

Now, the $\sqrt{-49}$ must be respresented as an imaginary number, and using the multiplication of radicals, we can simplify it to $\sqrt{49} \sqrt{-1}$
This means that we get the result 7i for the numerator.

$\frac{7i}{3-11i}$

In order to rationalize this fraction even further, we have to remember an identity from the previous algebra classes, namely: $x^2 - y^2 =(x+y)(x-y)$
The difference of squares allows us to remove the imaginary part of this fraction, leaving us with a real number, hopefully, on the denominator.

$\frac{7i (3+11i)}{(3-11i)(3+11i)}$

See, all I did there was multiply both numerator and denominator with (3+11i) so I could complete the difference of squares.
See how $(3-11i)(3+11i)= 3^2 -(11i)^2$ therefore, we can finally write:

$\frac{7i(3+11i)}{3^2 - (11i)^2 }$

I'll let you take it from here, all you have to do is simplify it further.
The simplification is quite straightforward, the numerator distributed the 7i. Namely the product $7i(3+11i) = 21i+77i^2$.
You should know from your classes that i^2 = -1, thefore the numerator simplifies to $-77+21i$
You can do it as a curious thing, but simplifying yields the result:
$\frac{-77+21i}{130}$