×

Help: Algebra

$\large \dfrac1{2+\sqrt3} = 2 -\sqrt3$

Why is the equation above true?

Note by Jason Chrysoprase
9 months, 4 weeks ago

Sort by:

$$\dfrac{1}{n+\sqrt{n+1}}=\dfrac{n-\sqrt{n+1}}{(n+\sqrt{n+1})(n-\sqrt{n+1})}=\dfrac{n-\sqrt{n+1}}{n^2-n-1}$$

We want denominator to be 1 so,

$$\implies n^2-n-1=1\\ \implies n^2-n-2=0 \\ \implies n^2-2n+n-2=0 \\ \implies n(n-2)+1(n-2)=0 \\ \implies (n+1)(n-2)=0$$

So it only works if $$n=-1,2$$ :) · 9 months, 4 weeks ago

You are smart, thx man · 9 months, 4 weeks ago

Write $$2$$ as $$\sqrt{4}$$,we'll get $\frac{1}{2+\sqrt{3}}=\frac{1}{\sqrt{3}+\sqrt{4}}=\frac{4-3}{\sqrt{4}+\sqrt{3}}=\frac{(\sqrt{4})^2-(\sqrt{3})^2}{\sqrt{4}+\sqrt{3}}=\frac{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}{\sqrt{4}+\sqrt{3}}=\sqrt{4}-\sqrt{3}=\boxed{2-\sqrt{3}}$ You could generalized to $$\frac{1}{\sqrt{n}+\sqrt{n+1}}$$ though · 8 months, 4 weeks ago

Nice ;) · 8 months, 4 weeks ago