Posted by **Anonymous** on Wednesday, September 29, 2010 at 11:09pm.

Is this how you solve t(t+1)(t+2)=0, x^3+4x=4x^2, and 3r^2=4r-1?

t(t+1)(t+2)=0

t=0 or t= -1 or t=2

{0, -1, -2}

x^3+4x=4x^2

x^3+4x-4x^2=0

x(x^2+4-4x)=0

x(x^2-4x+4)=0

x(x-2)(x-2)=0

x=0 or x=2

{0,2(double root)}

3r^2=4r-1

3r^2-4r+1=0

r^2-4r+3=0

(r-3)(r-1)=0

r=3 or r=1

{1,3}

- Algebra 2 -
**PsyDAG**, Thursday, September 30, 2010 at 1:32pm
The first two are fine.

The third one: 3r^2 - 4 + 1 = (3r-1)(r-1)

Take it from there.

- Algebra 2 -
**anonymous**, Thursday, September 30, 2010 at 7:26pm
thanks

it was {1/3, 1} right

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