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Express $\dfrac{1}{2+ \sqrt3}$ in its simplest form.

NB: The textbook has the answer as $2 - \sqrt3$ but I can't see how that was achieved.

I tried $\dfrac{1}{2} + \dfrac{1}{\sqrt3}$ and multiplying the top and bottom by $\sqrt3 $ to get $\dfrac{1}{2} + \dfrac{\sqrt3}{3}$ so far.

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In general, $\dfrac{1}{a+b} \neq \dfrac{1}{a} + \dfrac{1}{b}.$ –  Brad Jun 2 at 14:10
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For example, notice that $$\frac1{2+2}\ne \frac12+\frac12.$$ Similarly, it is also the case that $$\frac1{2+\sqrt3}\ne \frac12 + \frac1{\sqrt3}.$$ –  MJD Jun 2 at 14:11
    
How you split the fraction is a classic and very wrong move. –  orion Jun 2 at 14:11

2 Answers 2

Multiply by $\displaystyle 1 = \frac{2-\sqrt 3}{2-\sqrt 3}$. Use $(a-b)(a+b) = a^2-b^2$ on the denominator. It's called "rationalising the denominator" by multiplying the denominator by the "conjugate surd". You should look up the key phrases in quotes.

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Any time you are simplifying an expression like $$\frac{c}{a \pm \sqrt{b}},$$ multiply it with $$\frac{a\mp \sqrt{b}}{a\mp \sqrt{b}}$$ which gives you $$\frac{c(a\mp \sqrt{b})}{(a\pm \sqrt{b})(a\mp \sqrt{b})} = \frac{c(a\mp \sqrt{b})}{a^2 - b}$$ which has no more square roots in the denominator.

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