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Help: Algebra

\[ \large \dfrac1{2+\sqrt3} = 2 -\sqrt3 \]

Why is the equation above true?

Note by Jason Chrysoprase
1 year, 6 months ago

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

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 \(n=-1,2\) :)

Nihar Mahajan - 1 year, 6 months ago

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You are smart, thx man

Jason Chrysoprase - 1 year, 6 months ago

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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

Pham Khanh - 1 year, 5 months ago

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Nice ;)

Jason Chrysoprase - 1 year, 5 months ago

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Rationalise the denominator

Rushabh Zambad - 1 year, 6 months ago

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