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OpenStudy (anonymous):

Find the roots. 3x-2=3/-2x.

OpenStudy (anonymous):

First set the equation equal to f(x). The root of the function is where f(x) = 0.

OpenStudy (anonymous):

Is the equation 3x-2 = 3/-2(x) or 3x-2 = 3/(-2x)?

OpenStudy (anonymous):

The equation is 3x-2=(3/-2x). The number after the = sign is a fraction.

OpenStudy (anonymous):

right but is -2x on the bottom of the fraction or is it 3/-2 * x

OpenStudy (anonymous):

yes, -2x is on the bottom

OpenStudy (anonymous):

ok so f(x) = -6x^2 +4x -3 so to find the roots solve for x when -6x^2 + 4x -3 = 0

OpenStudy (anonymous):

alright, so now my question is how did you get the fraction to the other side of the eqution?

OpenStudy (anonymous):

by multiplying everything by (-2x) (3x-2)(-2x) = (3/-2x)(-2x) which simplifies to -6x^2 + 4x = -6x/-2x which further simplifies -6x^2 + 4x = 3, then move the 3 over so you get -6x^2 + 4x - 3 = 0

OpenStudy (anonymous):

if on the left side of teh equation the (-2x) is being distributed, why isn't the answer on the left side -6x/4x^2?

OpenStudy (anonymous):

because (3x-2)(-2x) is the same as 3x(-2x) - 2(-2x) which is -6x^2 - (-4x) which is -6x^2 + 4x

OpenStudy (anonymous):

oops im sorry i meant on the right side of the equation

OpenStudy (anonymous):

ok, its (3/-2x)(-2x) which is the same as 3(-2x)(1/-2x) the (-2x)(1/-2x) cancels out leaving only the 3

OpenStudy (anonymous):

im sorry im asking so many questions but where did you get the (1/-2x) from?

OpenStudy (anonymous):

3/-2x is the same as (3/1)* (1/-2x)

OpenStudy (anonymous):

ohh, okay that makes sense. thank you!

OpenStudy (anonymous):

no problem, I would use the quadratic formula to find the roots

OpenStudy (anonymous):

and luckily, that i remember how to do haha :)

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