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solve x^3 + 20x = -2x^3 - 7x 3x^3 + 27x = 0 3x (x^2 + 9) = 0 3x (x - 3) (x + 3) = 0 x = {-3, 0, 3} describe and correct the error by step please :)
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there is a miss take..... 3x(x^2+9)=0 ...... goood but 3x(x-3)(x+3)=0 is wrong
it should be 3x(x^2+3^2)=0
your final answer is right
how can someone solve it exactly if it were another type of problem?
x^2+9=0
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x^2=-9 x=3i
x=+- 3i
3x (x - 3) (x + 3) = 0 3x[x^2+3x-3x-9]=0 3x(x^2-9)=0 but you have 3x(x^2+9)
you just saved my life! xD thank youuuuu
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