How would you solve the equality:
$$a\,b\,\left(b+a\right)=1$$
in terms of a and/or b?
Would you subtract 1 from both sides and work from there? Or would you simply expand and work from there?
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Brian Fitzpatrick
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user134137
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You could solve a quadratic in $a$ or $b$ (the equation is symmetrical in $a,b$). – Pedro Mar 10 '14 at 07:03
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Quadratic Formula sounds good. Treat separately the case $a=0$. – André Nicolas Mar 10 '14 at 07:08
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@PedroTamaroff Thank you, I didn't spot that! – user134137 Mar 10 '14 at 07:09
2 Answers
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For the sake of closure I'll "answer" Pedro's answer in the comments. we have $ab^2+a^2b-1=0$, so we can use the quadratic equation to solve for one in terms of the other. This yields
$$b=\frac{-a^\frac{3}{2}\pm\sqrt{a^2+4}}{2\sqrt{a}}$$
Stella Biderman
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There's a typo in your solution (it should be $a^2+4$) and you've been missing one branch of the solution. – Michael Hoppe Mar 10 '14 at 11:56
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Let $b=ta$ for $t\notin\{-1,0\}$. Then $1=a^3t(t+1)$, that is $a=1/\sqrt[3]{t(t+t)}$ and $b=t/\sqrt[3]{t(t+1)}$.
Michael Hoppe
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