#How many of the solutions of the following equation are negative?

15 messages · Page 1 of 1 (latest)

nimble cedarBOT
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ancient turret
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Hm... Descartes' rule of signs gives 0 or 2 negative roots, I believe.
Maybe it's worth to find the derivative of f(x) = x^4 - 3x^3 - 6x^2 - 5x + 9 and see what happens to the function for x < 0?

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||Yeah, I believe for x < 0 f'(x) has a local (and absolute) maximum at x = -1/2 and is negative there, which implies that f'(x) < 0 for x < 0. But as f(0) > 0, that means that f(x) > 0 for x < 0, and thus doesn't have any negative roots.||

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||That does imply there are at least two positive roots, but even if we check many negative points, how can we be sure we don't "miss" the points that are good to use with IVT?||

ancient turret
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||Wait, two negative roots? But there aren't any negative roots.||

livid minnow
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I put 3x^2 instead of 3x^3

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Like a dummy

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You can apply Sturm’s theorem, this is a possibility, don’t know if it’s overkill though

ancient turret
stark hingeBOT
dapper shoal
unique flameBOT
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@supple laurel

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supple laurel
# livid minnow You can apply Sturm’s theorem, this is a possibility, don’t know if it’s overkil...

sorry i did not know the Sturm's theorem but i tried to solve it with Rational Root Theorem x^4−3x^3−6x^2−5x+9=0. , so i just calculated all possible negative solutions like -1 -3 and -9 but any of these possible solutions did not work, i tried to solve it with Sturm's theorem and it seems like further calculations will be too complex so i think there is a mistake in this equation. Thank you guys for your tips i will use them in my further calculations!