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Mathematics 13 Online
OpenStudy (anonymous):

Solve 4y^3-36y=0

OpenStudy (mertsj):

First factor you 4y

OpenStudy (mertsj):

factor out 4y I meant to type.

OpenStudy (anonymous):

(2y-6)(2y+6)=0

OpenStudy (anonymous):

y(2y-6)(2y+6)=0

OpenStudy (callisto):

1. factorise the equation on the left. 4y^3-36y=0 4y (y^2 -9) =0 4y (y-3)(y+3) =0 2. put each factor =0 4y=0 (y-3) =0 (y+3) =0 3. Solve them :)

OpenStudy (mertsj):

No @cinar

OpenStudy (anonymous):

4y(y^2)-9 = 0 4y(y+3)(y-3) y=0 OR y=-3 OR y=3

OpenStudy (anonymous):

why not..

OpenStudy (anonymous):

It's right @cinar

OpenStudy (anonymous):

can someone solve 4x^3-x=0

OpenStudy (anonymous):

x(4x^2 - 1)= 0 x = 0 or 4x^2 -1=0 (2x-1)(2x+1)=0 x= -1/2 OR x = 1/2 So the answer is : x=0 x= -1/2 x= 1/2

OpenStudy (anonymous):

@yarely

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