find zeros for y=(3x/(x+2)) + (5/(3x -1))

how would i go about doing this?

i set y=0 and got 0=9x^2 +2x +10/3x^2 -7x -2

if this is right i'm not sure what to do from here

thanks

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- Aug 25th 2008, 04:23 PMapmfind zeros for y=(3x/(x+2)) + (5/(3x -1))
find zeros for y=(3x/(x+2)) + (5/(3x -1))

how would i go about doing this?

i set y=0 and got 0=9x^2 +2x +10/3x^2 -7x -2

if this is right i'm not sure what to do from here

thanks - Aug 25th 2008, 04:53 PMskeeter
$\displaystyle \frac{3x}{x+2} + \frac{5}{3x-1} = 0$

$\displaystyle \frac{3x}{x+2} = \frac{-5}{3x-1}$

$\displaystyle (3x)(3x-1) = (-5)(x+2)$

$\displaystyle 9x^2 + 2x + 10 = 0$

since $\displaystyle b^2-4ac < 0$, there are no real zeros, i.e no solutions. - Aug 25th 2008, 04:56 PMJhevon