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Thread: Solve for 1/x

  1. #1
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    Solve for 1/x

    Question 18



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    Last edited by topsquark; Jul 12th 2018 at 12:48 PM.
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  2. #2
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    Re: Help #2

    Please use a descriptive title for your thread. It helps us (and you for that matter) to keep track of the conversations. "Help #2" tells us nothing.

    -Dan
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  3. #3
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    Re: Solve for 1/x

    Study this partial solution, HERE
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  4. #4
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    Re: Solve for 1/x

    Quote Originally Posted by sbjsavio View Post
    Question 18



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    Stuck here


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  5. #5
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    Re: Solve for 1/x

    Quote Originally Posted by sbjsavio View Post

    Stuck here
    Did you study this HERE ?
    If not then you are lazy.
    If you did you know that $\large x=\dfrac{b}{a}-1=\dfrac{b-a}{a},~a\ne 0$ so $\large\dfrac{4}{x}=\dfrac{4a}{b-a}$

    You can show that $\large 1+\dfrac{4}{x}+\dfrac{4}{x^2}=?$

    the answer is (2) HOW???
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  6. #6
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    Re: Solve for 1/x

    Quote Originally Posted by Plato View Post
    Did you study this HERE ?
    If not then you are lazy.
    If you did you know that $\large x=\dfrac{b}{a}-1=\dfrac{b-a}{a},~a\ne 0$ so $\large\dfrac{4}{x}=\dfrac{4a}{b-a}$

    You can show that $\large 1+\dfrac{4}{x}+\dfrac{4}{x^2}=?$

    the answer is (2) HOW???
    I end up getting a-b whole cube divided by ab
    And that wolfram alpha is not free,i need to buy premium to view it(not lazy)
    I didnt get that how u got x equals
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  7. #7
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    Re: Solve for 1/x

    $$abx^2=(a-b)^2(x+1)$$

    This is a quadratic in $x$:

    $$(ab)x^2 - (a-b)^2 x - (a-b)^2 = 0$$

    $$\begin{align*}x & = \dfrac{(a-b)^2 \pm \sqrt{(a-b)^4 + 4ab(a-b)^2}}{2ab} \\ & = \dfrac{(a-b)^2 \pm \sqrt{(a-b)^2[(a-b)^2+4ab]}}{2ab} \\ & = \dfrac{(a-b)^2 \pm \sqrt{(a-b)^2(a+b)^2}}{2ab} \\ & = \dfrac{(a-b)^2 \pm |a-b||a+b|}{2ab} \\ & = \dfrac{(a-b)^2 \pm |a^2-b^2|}{2ab}\\ & = \dfrac{(a-b)^2 \pm (a^2-b^2)}{2ab}\end{align*}$$

    Can you finish from there?
    Last edited by SlipEternal; Jul 13th 2018 at 05:42 AM.
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  8. #8
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    Re: Solve for 1/x

    rewrite the equation in the form

    $\displaystyle \frac{x+1}{x^2}=\frac{a b}{(a-b)^2}$

    $\displaystyle \frac{1}{x^2}+\frac{1}{x}=\frac{a b}{(a-b)^2}$
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  9. #9
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    Re: Solve for 1/x

    Ideas method is easiest.thanks man
    I got the answer
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