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anonymous
 one year ago
if x+1/x=7, what is x^3+1/x^2?
anonymous
 one year ago
if x+1/x=7, what is x^3+1/x^2?

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Is it \(\large\rm \frac{x+1}{x}=7\) or \(\large\rm x+\frac{1}{x}=7\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(x) + (1/x) = 7, the second one

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Fractions 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? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2There 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.2No? Hmm. Ok lemme make sure I'm understanding the second equation then, is it \(\large\rm x^3+\frac{1}{x^2}\) ?

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0\[x+\frac{1}{x}=7\] \[x=7\frac{1}{x}\] \[\frac{1}{x}=7x\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.2Woops typo,\[\large\rm x^2+\frac{1}{x^2}=47\]

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0\[x^3+\frac{1}{x^2}=(7\frac{1}{x})^3+\frac{1}{(7x)^2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3hero 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
 one year ago
Best ResponseYou've already chosen the best response.3\[(x+\frac{1}{x})^2=7^2 \\ (x+\frac{1}{x})^3=7^3\] some expanding will be involved

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you 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
 one year ago
Best ResponseYou've already chosen the best response.0seems like there is a more prosaic way of doing this

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you know what I think I want to reexpress 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
 one year ago
Best ResponseYou've already chosen the best response.0first 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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