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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$\displaystyle \frac{3x}{x+2} + \frac{5}{3x-1} = 0$ $\displaystyle \frac{3x}{x+2} = \frac{-5}{3x-1}$ $\displaystyle (3x)(3x-1) = (-5)(x+2)$ $\displaystyle 9x^2 + 2x + 10 = 0$ since $\displaystyle b^2-4ac < 0$, there are no real zeros, i.e no solutions.
Originally Posted by apm 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 your denominator is wrong, it wont affect the problem though. proceed by setting the numerator to zero and solving. as long as the denominator is not zero for the same values, those are the zeros of your function EDIT: geez, I am so slow today!
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