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Math Help - If [x+(1/x)]=3, find the value of x^5+[1/(x^5)].

  1. #1
    Junior Member Kaloda's Avatar
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    Question If [x+(1/x)]=3, find the value of x^5+[1/(x^5)].

    If (x+1/x)=3, then find the numerical value of x^5+1/x^5.
    Last edited by mr fantastic; September 21st 2010 at 12:44 AM. Reason: Re-titled.
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    Is it \frac{x + 1}{x} = 3 or x + \frac{1}{x} = 3?

    Are you trying to find \frac{x^5 + 1}{x^5} or x^5 + \frac{1}{x^5}?
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  3. #3
    Junior Member Kaloda's Avatar
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    I think you will understand this better:
    If [x+(1/x)]=3, then find the numerical value of x^5+[1/(x^5)].
    Last edited by Kaloda; September 20th 2010 at 11:31 PM.
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  4. #4
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    Quote Originally Posted by Kaloda View Post
    If (x+1/x)=3, then find the numerical value of x^5+1/x^5.
    Things to note:

    \displaystyle \left(x + \frac{1}{x}\right)^2 = 9 \Rightarrow x^2 + 2 + \frac{1}{x^2} = 9 \Rightarrow x^2 + \frac{1}{x^2} = 7 .... (A)

    From (A):

    \displaystyle \left(x + \frac{1}{x}\right) \left(x^2 + \frac{1}{x^2}\right) = (3)(7) = 21 .... (B)


    \displaystyle \left(x + \frac{1}{x}\right) \left(x^2 + \frac{1}{x^2}\right) = x^3 + x + \frac{1}{x} + \frac{1}{x^3} = \left(x^3 + \frac{1}{x^3}\right) + \left(x + \frac{1}{x}\right) \displaystyle = x^3 + \frac{1}{x^3} + 3 .... (C)

    From (B) and (C):

     \displaystyle x^3 + \frac{1}{x^3} = .... .... (D)

    etc.

    It's not hard to work your way to the value of \displaystyle x^5 + \frac{1}{x^5}.
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    Junior Member Kaloda's Avatar
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    Red face

    Thank you very very much!
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    \displaystyle{{x}^{3}}+\frac{1}{{{x}^{3}}}=\left( x+\frac{1}{x} \right)\left( {{\left( x+\frac{1}{x} \right)}^{2}}-3 \right)\text{ (1)}, and \displaystyle{{x}^{4}}+\frac{1}{{{x}^{4}}}={{\left  ( {{\left( x+\frac{1}{x} \right)}^{2}}-2 \right)}^{2}}-2\text{ (2)}\text{,} so \displaystyle\left( x+\frac{1}{x} \right)\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)={{x}^{5}}+\frac{1}{{{x}^{3}}}+{{x}^{3}}+\f  rac{1}{{{x}^{5}}}, conclude by (1) and (2).
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  7. #7
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    Hello, Kaloda!

    This solution should surprise/amaze/terrify your teacher . . .



    \text{If }x+\dfrac{1}{x} \:=\:3\text{, find the value of }x^5 + \dfrac{1}{x^5}

    We have: . x + \dfrac{1}{x} \:=\:3 .[1]

    Square both sides:
    . . \left(x + \dfrac{1}{x}\right)^2 \:=\:3^2 \quad\Rightarrow\quad x^2 + 2 + \dfrac{1}{x^2} \:=\:9 \quad\Rightarrow\quad x^2 + \dfrac{1}{x^2} \:=\:7



    Raise [1] to the 5th power: . \left(x + \dfrac{1}{x}\right)^5 \;=\;3^5


    . . . . . . . . . . . . \displaystyle x^5 + 5x^3 + 10x + \frac{10}{x} + \frac{5}{x^3} + \frac{1}{x^5} \;\;=\;\;243


    . . . . . \displaystyle \left(x^5 + \frac{1}{x^5}\right) + 5\underbrace{\left(x^3 + \frac{1}{x^3}\right)}_{\text{sum of cubes}} + 10\underbrace{\left(x + \frac{1}{x}\right)}_{\text{This is 3}} \;\;=\;\;243


    . . \displaystyle \left(x^5 + \frac{1}{x^5}\right) + 5\underbrace{\left(x + \frac{1}{x}\right)}_{\text{This is 3}} \underbrace{\left(x^2 - 1 + \frac{1}{x^2}\right)}_{\text{This is 7 - 1}} + 30 \;\;=\;\;243


    . . . . . . . . . . . . . . . . \displaystyle \left(x^5+\frac{1}{x^5}\right) + 5(3)(6) + 30 \;\;=\;\;243


    . . . . . . . . . . . . . . . . . . . Therefore: . . \displaystyle x^5 + \frac{1}{x^5} \;\;=\;\;123
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  8. #8
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    Quote Originally Posted by Kaloda View Post
    If (x+1/x)=3, then find the numerical value of x^5+1/x^5.
    You could also use the Binomial Expansion.

    \displaystyle\left(x+\frac{1}{x}\right)^5=x^5+\bin  om{5}{1}x^4\ \frac{1}{x}+\binom{5}{2}x^3\ \frac{1}{x^2}+\binom{5}{3}x^2\ \frac{1}{x^3}+\binom{5}{4}x\ \frac{1}{x^4}+\frac{1}{x^5}

    =\displaystyle\ x^5+5x^3+10x+10\ \frac{1}{x}+5\ \frac{1}{x^3}+\frac{1}{x^5}

    \displaystyle\Rightarrow\ \left(x+\frac{1}{x}\right)^5=x^5+\frac{1}{x^5}+10\  left(x+\frac{1}{x}\right)+5\left(x^3+\frac{1}{x^3}  \right)

    \displaystyle\Rightarrow\ x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-10\left(x+\frac{1}{x}\right)-5\left(x^3+\frac{1}{x^3}\right)


    Also

    \displaystyle\left(x+\frac{1}{x}\right)^3=x^3+\bin  om{3}{1}x^2\ \frac{1}{x}+\binom{3}{2}x\ \frac{1}{x^2}+\frac{1}{x^3}

    =\displaystyle\ x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)

    \displaystyle\Rightarrow\ x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)


    Therefore

    \displaystyle\ x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-10\left(x+\frac{1}{x}\right)-5\left[\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right]

    \displaystyle\Rightarrow\ x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(x+\frac{1}{x}\right)^3+5\left(x+\frac{1}{x}  \right)

    =3^5-(5)3^3+5(3)
    Last edited by Archie Meade; September 22nd 2010 at 02:32 AM. Reason: tidy up
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  9. #9
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    x^5 + 1/x^5 = 123

    x = {[123+-SQRT(15125)] / 2}^(1/5)

    x = 2.618033988.... or .381966011.... ; ya, ya, so what!
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