TeemoTheTerific
help me on a question please...
x= 1/2 +1/2sqrt5
show that...
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TeemoTheTerific
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\[x=\frac{ 1 }{ 2 } + \frac{ 1 }{ 2 }\sqrt{5}\]
TeemoTheTerific
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\[\frac{ x^{2}+x ^{-2} }{ x-x ^{-1} }=3\]
Azteck
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\[x=\frac{1}{2}+\frac{\sqrt{5}}{2}\]
\[x=\dfrac{1+\sqrt{5}}{2}\]
enka
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\[x=\frac{1+\sqrt{5}}{2}\\
\frac{1}{x}=\frac{2}{1+\sqrt{5}}\times\frac{\sqrt{5}-1}{\sqrt{5}-1}=\frac{2\sqrt{5}-1}{4}=\frac{\sqrt{5}-1}{2}\]
then
\[x-\frac{1}{x}=1\qquad(1)\]
square both side to get:
\[x^2+\frac{1}{x^2}-2=1\\
x^2+\frac{1}{x^2}=3\qquad(3)\]
now divide (3) by (1) to get the result
Azteck
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\[x^2+x^{-2}=3x-3x^{-1}\]
Multiply both sides by x^2.
\[x^4+1=3x^3-3x\]
\[x^4-3x^3+3x+1=0\]
TeemoTheTerific
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i dont get it... both of you :P
TeemoTheTerific
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yes ok
enka
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first, we find the value of \[x-\frac{1}{x}\]
Azteck
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You just sub in \[\frac{1+\sqrt{5}}{2}\]
into the original equation....
Azteck
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And you prove that the LHS=RHS.
TeemoTheTerific
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but no matter how manytimes i try i can never get it to equal 3
Azteck
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Use a calculator if you can't do it....Use a calculator to solve the denominator first. You get one as the denominator...
Azteck
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Jeez, you have a tool that can help you but you don't want to use it.
Azteck
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You get 3 as the denominator when you plug the values into the numerator...
Azteck
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numerator*
TeemoTheTerific
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we need to show our working our using our skill of indices laws... But i dont have much
Azteck
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There's no indice work when you learn that
\[x^{-1}=\frac{1}{x^1}\]
\[x^{-2}=\frac{1}{x^2}\]
Azteck
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Dude, now once you see the letter x, put that big fat fraction into your calculator and get your values...
Azteck
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I can solve that by hand but that will take me ages, to write it on the computer, so no thank you...
enka
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i hope it is can help you
enka
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or this
TeemoTheTerific
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thank you enka
enka
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you're welcome :)