# 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?

• How did you compute $4$ in the denominator? Jan 1 '19 at 18:15
• @MartinR my train of thought was 2 * 2 and then that $+\sqrt{3}$ cancels out $-\sqrt{3}$ Jan 1 '19 at 18:18
• Why the downvote??? Jan 1 '19 at 18:22

Note that $$\left(2+\sqrt3\right)\times\left(2-\sqrt3\right)=4-3=1$$.

• 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? Jan 1 '19 at 17:57
• @DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$. Jan 1 '19 at 17:58

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 $$a^2-b^2=4-3=1$$

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!

• Thank you for the tip! Jan 1 '19 at 20:32