anonymous
  • anonymous
help me on a question please... x= 1/2 +1/2sqrt5 show that...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[x=\frac{ 1 }{ 2 } + \frac{ 1 }{ 2 }\sqrt{5}\]
anonymous
  • anonymous
\[\frac{ x^{2}+x ^{-2} }{ x-x ^{-1} }=3\]
anonymous
  • anonymous
\[x=\frac{1}{2}+\frac{\sqrt{5}}{2}\] \[x=\dfrac{1+\sqrt{5}}{2}\]

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

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