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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Jun 2007 15:21:33 IST
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find the no. of real root of the polynomial f(x) = x^5+x^3-2x+1
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1 Jun 2007 15:24:20 IST
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there are 3real roots rate me if useful
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1 Jun 2007 16:31:57 IST
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f(x) = x5 + x3 - 2x + 1
f'(x) = 5x4 + 3x2 - 2 = (5x2-2)(x2+1)
f'(x) = 0 has roots x = 2/ 5 , -2/ 5
i.e. f(x) has maxima at x = 2/5 nad minima at x = -2/5.
f(2/5) < 0
f(-2/5) > 0
hence maxima lies above the x-axis and minima lies below the x-axis.
Hence curve intersects x-axis in between minimal and maximal points.
Hence curve intersects x-axis at 3 points one before maxima and one after minima and in between.
So 3 real solutions possible.
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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