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Solve for x x^2(x-3)^2-4(x-3)^2=0
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\[x^2(x-3)^2-4(x-3)^2=0\] Factoring out (x - 3)², you have \[(x-3)^2(x^2-4)=0\]
ok got that\
so then you get (x^2-6x+9)x(x-4) right
No, that doesn't help. What I said tells you that you have either \[(x-3)^2=0 \text{ or }x^2-4=0.\] Solving both for x gives you the solutions.
I am stuck for the x(x-4) I got htat far what next
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Is it x=3 and x=-4
Can someone please help me
Sorry, was a bit busy. You're right about the x = 3 as a solution, but not with the other equation. \[x^2-4=0\\ x^2=4\\ x=\pm\sqrt{4}\\ x=-2,x=2\]
so there are three answers x=-2 and x=2 and x=3
Yes.
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thanks
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