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Min Hoang moderators

16/03/2017 at 10:34
Answers
2
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Solve for x:

\(\dfrac{1}{x^2+18x-20}+\dfrac{1}{x^2+18x-24}=\dfrac{-12}{x^2+18x-29}\)


Equation


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    mathlove 16/03/2017 at 11:33

    Put  \(t=x^2+18x-24\Leftrightarrow x^2+18x-24-t=0\Leftrightarrow x=-9\pm\sqrt{105+t},\left(t\ge-105\right)\), the equation become    

                                                      \(\dfrac{1}{t+4}+\dfrac{1}{t}=\dfrac{-12}{t-5}\)  (1)

    With the condition  \(t\notin\left\{0;-4;5\right\}\), \(\left(1\right)\Leftrightarrow7t^2+21t-10=0\Leftrightarrow t=\dfrac{-21\pm\sqrt{721}}{14}\)(both of these solutions are not smaller  -105). So, the done equation have 4 solutions

                     \(x=-9\pm\pm\sqrt{\dfrac{1449+\sqrt{721}}{14}};x=-9\pm\sqrt{\dfrac{1449-\sqrt{721}}{14}}\)  .

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    FA KAKALOTS 28/01/2018 at 22:11

    put  t=x2+18x−24⇔x2+18x−24−t=0⇔x=−9±√105+t,(t≥−105)

    , the equation become    

                                                      1t+4+1t=−12t−5

      (1)

    With the condition  t∉{0;−4;5}

    , (1)⇔7t2+21t−10=0⇔t=−21±√72114

    (both of these solutions are not smaller  -105). So, the done equation have 4 solutions

                     x=−9±±√1449+√72114;x=−9±√1449−√72114

      .


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