$$x^5-1102x^4-2015=0$$
Would really appreciate if you could tell me in detail how to solve this question. I only know I should differentiate it. I don't know how to continue from there on.
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Could it be that you misquoted the task and it is really a duplicate of http://math.stackexchange.com/q/1439912/115115 ? I.e., it is $x^5−1102x^4−2015x=0$? – Lutz Lehmann Sep 24 '15 at 12:32
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Your question, as posted, is incorrect. There is only a single real root, and it is a positive one.
Proving it can be done using the Decartes' Rule of Signs: http://www.purplemath.com/modules/drofsign.htm
There is just one sign change here giving exactly one positive real root. If you substitute $-x$ for $x$ you can show via the absence of sign changes in the transformed equation that there are no negative real roots.
Finding the single positive real root approximately is not hard given that the first two terms will quickly overwhelm the constant term, so the only positive real solution will be very close to $1102$. In fact, it's something like $1102.000000002$
All the other $4$ roots are complex, and they come in two pairs of conjugate complex numbers.
Deepak
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One possible solution - one in which you have to differentiate - is applying Fourier's theorem. Essentially, what you have to do is to differentiate the polynomial five times, obtaining a sequence of polynomials. Then you evaluate all derivatives at $\infty$, and count the number of sign changes in the resulting sequence. Repeat this for $-\infty$ and compare the number of sign changes to the number you obtained for $\infty$. The difference equals the number of real roots.
Bernhard
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There are four complex solutions, near the solutions of $-1102x^4-2015=0$, and one real solution, near the large solution of $x^5-1102x^4=0$
Empy2
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