anonymous
  • anonymous
if x+1/x=7, what is x^3+1/x^2?
Mathematics
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SOLVED
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katieb
  • katieb
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zepdrix
  • zepdrix
Is it \(\large\rm \frac{x+1}{x}=7\) or \(\large\rm x+\frac{1}{x}=7\) ?
anonymous
  • anonymous
(x) + (1/x) = 7, the second one
zepdrix
  • zepdrix
Fractions are bad. No bueno. So let's multiply through to get rid of the x in the denominator. Multiplying by x gives us:\[\large\rm x^2+1=7x\]Ok with that step? What should we do next, what do you think? :)

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zepdrix
  • zepdrix
There is probably a cool way to relate the first equation to the second one... but I'm not seeing it. So instead what we can do is.. simply solve for the value of x in the first equation, and plug that value into the second equation.
anonymous
  • anonymous
I do not think that is the way my teacher would like for me to go about it. I have been staring at this problem for a good 35 minutes and I'm stumped.
zepdrix
  • zepdrix
No? Hmm. Ok lemme make sure I'm understanding the second equation then, is it \(\large\rm x^3+\frac{1}{x^2}\) ?
anonymous
  • anonymous
Yes it is.
Mertsj
  • Mertsj
\[x+\frac{1}{x}=7\] \[x=7-\frac{1}{x}\] \[\frac{1}{x}=7-x\]
zepdrix
  • zepdrix
I was thinking maybe we could do something like this:\[\large\rm x+\frac{1}{x}=7\]Then,\[\large\rm \left(x+\frac{1}{x}\right)^2=49\]and\[\large\rm x^2+\frac{1}{x}=47\]But that doesn't quite get us the third power :( Darn...
zepdrix
  • zepdrix
Woops typo,\[\large\rm x^2+\frac{1}{x^2}=47\]
Mertsj
  • Mertsj
\[x^3+\frac{1}{x^2}=(7-\frac{1}{x})^3+\frac{1}{(7-x)^2}\]
freckles
  • freckles
hero had a good way on a previous post like this one something about squaring both sides and also cubing both sides of that one equation
freckles
  • freckles
\[(x+\frac{1}{x})^2=7^2 \\ (x+\frac{1}{x})^3=7^3\] some expanding will be involved
freckles
  • freckles
you will get values for both \[x^2+\frac{1}{x^2} \text{ and } x^3+\frac{1}{x^3} \\ \text{ then add those you get } \\ x^3+\frac{1}{x^2}+x^2+\frac{1}{x^3} \\ =x^3+\frac{1}{x^2}+\frac{1}{x}(x^3+\frac{1}{x^2}) \\ =(x^3+\frac{1}{x^2})(1+\frac{1}{x}) \\ \text{ so you have } \\ x^3+\frac{1}{x^2}+x^2+\frac{1}{x^3}=(x^3+\frac{1}{x^2})(1+\frac{1}{x}) \\ \text{ solving for } x^3+\frac{1}{x^2} \\ \text{ you have } \\ x^3+\frac{1}{x^2}=\frac{x^3+\frac{1}{x^2}+x^2+\frac{1}{x^3}}{1+\frac{1}{x}}\] and remember you found the numerator above from doing what hero suggested and then you are given the denominator
anonymous
  • anonymous
seems like there is a more prosaic way of doing this
freckles
  • freckles
you know what I think I want to re-express what I said just a bit: \[x^3+\frac{1}{x^2}=\frac{(x^3+\frac{1}{x^3})+(x^2+\frac{1}{x^2})}{1+\frac{1}{x}}\] just wanted to group together what we actually will find from the above equations that I mentioned (or I mean hero had mentioned in a previous post)
anonymous
  • anonymous
first equation tells you \[x=\frac{7\pm3\sqrt{5}}{2}\] it is much less fun, but i am sure will get an answer

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