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x^3=3x^2+16 and i'm looking for x :)
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x^3=3x^2+16 x^3-3x^2-16=0
Now factor the above cubic polynomial..
can u?
not at all, can you do this ?
x^3=3x^2+16 x^3-3x^2-16=0 x^3-4x^2+x^2-4x+4x-16=0 x^2(x-4)+x(x-4)+4(x-4)=0 (x-4)(x^2+x+4)=0
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Can u continue from here?
from (x^2+x+4) is only delta?
ok, but delta <0, so x=4, right?
Yep..... x=4 is the only real solution.
thank you very much :)
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