anonymous
  • anonymous
if x+1/x =8, find the x^2 + x^-2
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Is it \[{x+1 \over x}=8?\]
anonymous
  • anonymous
no, it is x+(1/x)
anonymous
  • anonymous
(x + 1/x)^2 = x^2 + 2 + 1/x^2

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anonymous
  • anonymous
It is further = x^2 + 2 + x^-2
anonymous
  • anonymous
finally (x + 1/x)^2 - 2 = x^2 + x^-2 a
anonymous
  • anonymous
\[x+{1 \over x}=8 \implies x^2+2+x^{-2}=16 \implies x^2+x^{-2}=14\]
anonymous
  • anonymous
You hv x + 1/x = 8 Squaring both sides (x + 1/x)^2 = 8^2 = 16 S0, 16 - 2 = x^2 + x^-2 so 14 is the answer
anonymous
  • anonymous
Sorry not 16, it's 64. So the final answer is 62.
anonymous
  • anonymous
the answer I have are: a) 60, b)58, c)62, d)0, e)non So the answer must be e)
anonymous
  • anonymous
Correct Anwar!!!
anonymous
  • anonymous
So answer is 62
anonymous
  • anonymous
option c)
anonymous
  • anonymous
\[x+{1 \over x}=8 \implies (x+{1 \over x})^2=8^2 \implies x^2+2+x^{-2}=64 \] \[\implies x^2+x^{-2}=62\]
anonymous
  • anonymous
Here is the corrected version You hv x + 1/x = 8 Squaring both sides (x + 1/x)^2 = 8^2 = 64 S0, 64 - 2 = x^2 + x^-2 so 62 is the answer
anonymous
  • anonymous
You got the idea emunrradtvamg?
anonymous
  • anonymous
yes, thanks guys its clear now
anonymous
  • anonymous
You're welcome!

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