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

• Aug 25th 2008, 04:23 PM
apm
find 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 PM
skeeter
$\frac{3x}{x+2} + \frac{5}{3x-1} = 0$

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

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

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

since $b^2-4ac < 0$, there are no real zeros, i.e no solutions.
• Aug 25th 2008, 04:56 PM
Jhevon
Quote:

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!