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Question: $$\frac{5m}{2}=2+\frac{1}{m}$$

I have attempted the question but my answer is not correct according to the book. $$\frac{5m^2}{2m}-2=0$$ $$5m^2-2=2m$$ $$5m^2-2m-2=0$$ When I placed my following working out into the quadratic formula my answer was incorrect. Is my working out incorrect for the above? Help much appreciated. Thank you in advance.

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  • $\begingroup$ if you multiply both sides by $2m,$ you get $5m^2 = 4m + 2.$ your middle term should be $-4m$ not $-2m$ as you have it. $\endgroup$ – abel Feb 7 '15 at 9:23
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Multiply both sides by $m$

$\implies \frac{5m^2}{2}=2m+1$

$\iff 5m^2=4m+2$

$\iff 5m^2-4m-2=0$

Then you can factorize by using $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b-\sqrt{b^2-4ac}}{2a}$ or some other method that you prefer.

$\implies \frac{1}{5}(2-\sqrt{14})$ and $\frac{1}{5}(2+\sqrt{14})$

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You first step is incorrect. You took $1/m$ to LHS. Check what you did after it. There must be a additional $-2$ along with $5m^2$ in the denominator in the first step. Finally you should have the quadratic $5m^2-4m-2=0$

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