# 9. Prove that0 hasThe equation x+ ax+ b2one(a)negative and two imaginary roots if b 0.TL

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#9a using Descartes's rule of sign help_outlineImage Transcriptionclose9. Prove that 0 has The equation x+ ax+ b2 one (a) negative and two imaginary roots if b 0. TL fullscreen
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Step 1

Given equation is,

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, help_outlineImage Transcriptionclosen-(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 fullscreen
Step 3

Consider, b which is not equ... help_outlineImage Transcriptionclosexax2b0 . (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 fullscreen

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