SOLUTION: hey ya all thanx to your help I havent be around here for a while...but even the best of luck faids away sometimes, and I am back again, with problems again, would someone be as ni
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Question 199483: hey ya all thanx to your help I havent be around here for a while...but even the best of luck faids away sometimes, and I am back again, with problems again, would someone be as nice as to help me out here again? =) Please
My problem; Approximate to the nearest tenth, the real root of the equation f(x)=x^3-4=0, I dont at all get this one.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Finding the real root of this cubic equation is pretty straightforward.
Take the cube root of both sides:
Use your calculator and round to the nearest tenth.
But I suspect that you are bothered by the fact that the question specifies the real root which implies that there is something other than the real root available. Indeed there is.
Since this is a 3rd degree polynomial, the Fundamental Theorem of Algebra guarantees that there are exactly 3 factors of the form 
.
Note that 
can be considered the difference of two cubes to the extent that 4 is the cube of the cube root of 4.
The factorization of the difference of two cubes is:
 = (a - b)(a^2 + ab + b^2))
Which, in the case of your problem would look like:
 = (x - \sqrt[3]{4})(x^2 + \left(\sqrt[3]{4}\right)x + \sqrt[3]{4^2}) = (x - \sqrt[3]{4})(x^2 + \left(\sqrt[3]{4}\right)x + 2\sqrt[3]{2}))
The first factor yields the same real root that we calculated earlier. The second factor is a quadratic that is guaranteed to have two factors, but if you use the discriminant, you will see that the roots of that quadratic are a conjugate pair of complex numbers. Hence, the original given equation has one real and two complex roots.
John
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