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Math Help - (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

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    (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

    \left(x+\frac{1}{x}\right)^2-\frac{3}{2}\left(x-\frac{1}{x}\right)

    When expanding the square, and adding the fractions I get a quartic equation. Is there any way?
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    Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

    It may be a quartic equation in x, but it appears at first glance (without trying it out) to be a quadratic equation in x^2.
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    Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

    Or: let y= x+ \frac{1}{x}, solve the quadratic equation y^2- (3/2)y= 4 for y, then solve x+ \frac{1}{x}= y which will be another quadratic equation, for x. (The roots are NOT easy integer or even rational numbers.)
    Last edited by HallsofIvy; July 24th 2014 at 07:16 AM.
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    Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

    Hello, NameIsHidden!

    \left(x+\frac{1}{x}\right)^2-\frac{3}{2}\left(x-\frac{1}{x}\right)\;=\;4

    Watch what happens when we expand . . .

    \quad\left(x+\frac{1}{x}\right)^2 - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;4

    x^2 + 2 + \frac{1}{x^2} - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;4

    x^2 - 2 + \frac{1}{x^2} - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;0

    \left(x -\frac{1}{x}\right)^2 - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;0


    Let u \,=\,x+\frac{1}{x}\!:\;\;u^2 - \tfrac{3}{2}u \:=\:0 \quad\Rightarrow\quad u\left(u - \tfrac{3}{2}\right) \:=\:0

    Hence: . u \,=\,0,\;u \,=\,\tfrac{3}{2}


    Back-substitute:

    . . x + \frac{1}{x} \:=\:0 \quad\Rightarrow\quad x^2-1\:=\:0 \quad\Rightarrow\quad \boxed{x \:=\:\pm1}

    . . x + \frac{1}{x} \:=\:\frac{3}{2} \quad \Rightarrow\quad 2x^2 - 3x - 2 \:=\:0 \quad\Rightarrow\quad (x-2)(2x+1) \:=\:0

    . . . . . . \boxed{x \;=\;2.\:\text{-}\tfrac{1}{2}}

    Thanks from NameIsHidden
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    Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

    Quote Originally Posted by HallsofIvy View Post
    Or: let y= x+ \frac{1}{x}, solve the quadratic equation y^2- (3/2)y= 4 for y, then solve x+ \frac{1}{x}= y which will be another quadratic equation, for x. (The roots are NOT easy integer or even rational numbers.)
    Thank you everybody. HallsofIvy, you should have seen the negative sign as positive in -3/2(x-1/x)
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    Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

    Quote Originally Posted by Matt Westwood View Post
    It may be a quartic equation in x, but it appears at first glance (without trying it out) to be a quadratic equation in x^2.
    Sorry, ignore me, I was reading the second term as squared, and it appears it's not.
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    Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

    Quote Originally Posted by NameIsHidden View Post
    (x+1/x)^2-(3/2)(x-1/x)=4.
    Dollars to donuts you missed a couple pairs of brackets and that should be:
    [(x+1) / x]^2 - (3/2)[(x-1) / x] = 4.

    Then you'll get a "nice" quadratic: 9x^2 - 7x - 2 = 0
    and you'll happily end up with x = 1 or x = -2/9

    Yes? No?
    Thanks from topsquark
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    Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

    Quote Originally Posted by Wilmer View Post
    Dollars to donuts you missed a couple pairs of brackets and that should be:
    [(x+1) / x]^2 - (3/2)[(x-1) / x] = 4.

    Then you'll get a "nice" quadratic: 9x^2 - 7x - 2 = 0
    and you'll happily end up with x = 1 or x = -2/9

    Yes? No?
    No. You are wrong. I did not miss out any brackets. I gave all the brackets necessary. Soroban got it right.
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    Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

    Quote Originally Posted by HallsofIvy View Post
    Or: let y= x+ \frac{1}{x}, solve the quadratic equation y^2- (3/2)y= 4 for y, then solve x+ \frac{1}{x}= y which will be another quadratic equation, for x. (The roots are NOT easy integer or even rational numbers.)
    The equation has the term : -(3/2) (x-1/x) ,so if you put x +1/x=y ,can you have :x-1/x =y??
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    Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

    Quote Originally Posted by Soroban View Post
    Hello, NameIsHidden!


    Watch what happens when we expand . . .

    \quad\left(x+\frac{1}{x}\right)^2 - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;4

    x^2 + 2 + \frac{1}{x^2} - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;4

    x^2 - 2 + \frac{1}{x^2} - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;0

    \left(x -\frac{1}{x}\right)^2 - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;0


    Let u \,=\,x+\frac{1}{x}\!:\;\;u^2 - \tfrac{3}{2}u \:=\:0 \quad\Rightarrow\quad u\left(u - \tfrac{3}{2}\right) \:=\:0

    Hence: . u \,=\,0,\;u \,=\,\tfrac{3}{2}


    Back-substitute:

    . . x + \frac{1}{x} \:=\:0 \quad\Rightarrow\quad x^2-1\:=\:0 \quad\Rightarrow\quad \boxed{x \:=\:\pm1}

    . . x + \frac{1}{x} \:=\:\frac{3}{2} \quad \Rightarrow\quad 2x^2 - 3x - 2 \:=\:0 \quad\Rightarrow\quad (x-2)(2x+1) \:=\:0

    . . . . . . \boxed{x \;=\;2.\:\text{-}\tfrac{1}{2}}

    u=x-1/x. am I right?
    I think you made a typo once and did ctrl+c, ctrl+v. Anyway the answers are correct. Hence it must be a typo
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    Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

    Quote Originally Posted by NameIsHidden View Post
    u=x-1/x. am I right?
    I think you made a typo once and did ctrl+c, ctrl+v. Anyway the answers are correct. Hence it must be a typo
    Yes it is, because x+1/x does not imply x^2-1
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