logarithms please help !

Jan 2015
3
0
Asia
how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2?
 

topsquark

Math Team
May 2013
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The Astral plane
how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2?
\(\displaystyle \frac{1}{\sqrt{3} + \sqrt{2}}\)

Try multiplying the numerator and denominator by the conjugate:
\(\displaystyle \frac{1}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\)

Can you finish it?

-Dan
 
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Jan 2015
3
0
Asia
but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..?
 

skeeter

Math Team
Jul 2011
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but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..?
much easier the way topsquark has recommended ... and that is working "from the back"
 
Apr 2014
583
177
Europa
but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..?

\(\displaystyle \color{blue}{\sqrt3-\sqrt2=(\sqrt3-\sqrt2)\cdot \dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}= \dfrac{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}{\sqrt3+\sqrt2}=\dfrac{( \sqrt3)^2-(\sqrt2)^2}{\sqrt3+\sqrt2} = \dfrac{3-2}{\sqrt3+\sqrt2} = \dfrac{1}{\sqrt3+\sqrt2}}\)
 
Apr 2014
583
177
Europa
logarithms please help !

how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2?

Where is the logarithm ...?
 
Jan 2015
3
0
Asia
sorry! because this question was in my logarithm homework questions so I thought I needed to use logarithms! But thank you for your help !!
 
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