<s>[INST] Rationalizing denominator of $\frac{7}{2+\sqrt{3}}$. Cannot match textbook solution I am given this expression and asked to simplify by rationalizing the denominator:
$$\frac{7}{2+\sqrt{3}}$$
The solution is provided:
$14 - 7\sqrt{3}$
I was able to get to this in the numerator but am left with a 4 in the denominator. Here are my steps:
$$\frac{7}{2+\sqrt{3}} * \frac{2-\sqrt{3}}{2-\sqrt{3}}$$
=
$$\frac{14 - 7\sqrt{3}}{4}$$
Presumably I should not have a denominator here since the solution I'm given is just whats in the numerator. Presumably my numerator calculation is correct, where did I go wrong on the denominator?
 [/INST] Trick to remember forever and use again and again.
$(a + b)(a-b) = a(a-b) + b(a-b)=$
$a^2 - ab + ab - b^2 = a^2 - b^2$.
So
1) Whenever you need to factor  $a^2 - b^2$ it always factor to to $(a+b)(a-b)$
and 
2)  If you ever need to get rid of a radical sign in $a+\sqrt b$ you can always multiple by $(a + \sqrt b)(a - \sqrt b) = a^2 - \sqrt b^2 = a^2 -b$.
So 
3) to deradicalize a $\frac m{\sqrt a + \sqrt b} = \frac {m(\sqrt a - \sqrt b)}{(\sqrt a - \sqrt b)} = \frac {m(\sqrt a - \sqrt b)}{a - b}$.
So:
$\frac {7}{2 + \sqrt 3} = $
$\frac {7(2 - \sqrt 3)}{(2+\sqrt 3)(2 - \sqrt 3)} =$
$\frac {7(2-\sqrt 3)}{2^2 - \sqrt 3^2} =$
$\frac {7(2-\sqrt 3)}{4-3} =$
$\frac {7(2-\sqrt 3)}{1} =$
$7(2-\sqrt 3)$.
Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.
You will be using it for the REST OF YOUR LIFE!
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