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Asked Nov 24, 2019
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#9a using Descartes's rule of sign 

9. Prove that
0 has
The equation x+ ax+ b2
one
(a)
negative and two imaginary roots if b 0.
TL
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9. Prove that 0 has The equation x+ ax+ b2 one (a) negative and two imaginary roots if b 0. TL

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Expert Answer

Step 1

Given equation is,

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x3 axb0

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Step 2

Descartes Rule for positive roots: The number of positive real zeroes in a polynomial function  f (x) is the same or less than by an even number as the number of changes in the sign of the coefficients.

Descartes Rule for negative roots: The number of negative real roots is the number of sign changes after putting (– x) in place of x in f (x), or fewer than it by an even number.

 

Descartes Rule for imaginary roots: The number of imaginary roots is,

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n-(p q)where pis the number of positive roots of the polynomial, q is the number of negative roots of the polynomial and nis the highest degree of the polynomial

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Step 3

Consider, b which is not equ...

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xax2b0 . (1) In above equation there is no change in sign Hence, there is no positive root. Now, replace x by - x.in equation (1) -x3 axb 0 In above equation there is only one change in sign Hence, there is one negative root. Then, number of imaginary roots is, 3-(0+1) 3-1= 2 that is 2 imaginary roots

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