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=4r1?
t(t+1)(t+2)=0
t=0 or t= 1 or t=2
{0, 1, 2}
x^3+4x=4x^2
x^3+4x4x^2=0
x(x^2+44x)=0
x(x^24x+4)=0
x(x2)(x2)=0
x=0 or x=2
{0,2(double root)}
3r^2=4r1
3r^24r+1=0
r^24r+3=0
(r3)(r1)=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 = (3r1)(r1)
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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