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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?

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    $\begingroup$ How did you compute $4$ in the denominator? $\endgroup$ – Martin R Jan 1 at 18:15
  • $\begingroup$ @MartinR my train of thought was 2 * 2 and then that $+\sqrt{3}$ cancels out $-\sqrt{3}$ $\endgroup$ – Doug Fir Jan 1 at 18:18
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    $\begingroup$ Why the downvote??? $\endgroup$ – Randall Jan 1 at 18:22
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Note that $\left(2+\sqrt3\right)\times\left(2-\sqrt3\right)=4-3=1$.

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  • $\begingroup$ Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part? $\endgroup$ – Doug Fir Jan 1 at 17:57
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    $\begingroup$ @DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$. $\endgroup$ – Lord Shark the Unknown Jan 1 at 17:58
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note that: $$(a+b)(c+d)=ac+ad+bc+bd$$ so in some cases it simplifies to: $$(a+b)(a-b)=a^2-b^2$$ for you, $a=2$ and $b=\sqrt{3}$ so $a^2-b^2=4-3=1$

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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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  • $\begingroup$ Thank you for the tip! $\endgroup$ – Doug Fir Jan 1 at 20:32

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