Question

10.3 Descartes' Rule of Signs (a) If c. C2, ..., Cm are any m nonzero real numbers, and if 2 consecutive terms of this sequence have opposite signs, we say that these 2 terms present a variation of sign. With this concept, we may state Descartes' rule of signs, a proof of which may be found in any textbook on the theory of equations, as follows: Let f(x) = 0 be a polynomial equation

Answer 1

13 towa 10.3 W (1) 2° +32° -52°+ 4x +6 No. of sign changes in fix) ane f(x) = x + 3x - 5x² + 4x + 6 Чkhu Омо 2 пап лалын сав , Ілела are either 2 ou o posetuvre real roots. Now, in fl-2) fox) = Gaja +31-29_5(-x)² + 4 (-1) + 6 = -x9 +378 + sus UX +6 cu there are a sign In changes, so there are either 3 ou a negative real roots. So, possibilities are i- I Positive I Negative complen total real roots Real roots | roots งา م رو ย و 6 ه| (2) 2x² – 3x"_x3_5=0 fix) = 2x² – 3x4x²_5. there is only a sign change, so t positive real root. f(-x) = -247 24 so either a Possibilities are total there are 2 sign changes, or o negative real roots. Positive Negatiue I complen real rosts | real roots | roots I I r 2 l u adtanned with

(13) 3X" +10x2 +5X-4=0 flox) = 37" + lox² + 5x-4 + sign change. so there is only t. positive real root. H-2) = 324 +lox² -51.-4 . t sign change. so, there is only I negative real root so only possibility is that fall has t positive real, I negative ersal and 2 complen roots. aned with 3 Scanner

IZO f(x) = x^_1 yn is even frac) - X - only a positive real root t sign change means fl-1) = (-xnot in is even no ff-x) = x-1 t sign change means only to negative real root- so total nou of real roots = t1 = 3 Hence proved in is odd: cruz in (©) o sign change means t positive real root fl-x) = (-on-1 n is odd fl-x) = -xml no sign change, so o negative real roots 7 Wotal no. of real roots = 1 Henee proved x3+x²+1=0 f(x) = as tx²+1 no sign change means o positive real roots f-x) = -4++1 t sign change means to regature reel root. a only possibility is o positive real , t negative reed and ů imaginary for complex) ruots, AAA Hence proved (d) x² + parq=0 ; q, q are real - p = positive, a fla) = x² + parq qto CS Scanned with ÇamScanner

i qus positive then flod = x² + patq . no sign change means o positive real Hoots and fl-x) = x _ pt + q ( q us positure ) t sign change means i negative real root. & In this case , o positive roots, - negative root and 2 imaginary rests (total 3). Hence proved if q is negative : then fix) = x² + 2x +q (e q es negative ) to sign change means & positive real root f-) = x-parq no sign change means o negative real roots. al on this case, we have t positive real, o negative real and a imaginary boots (Motal 3) and - Hence proved, no etue M (e) we will use induction for the proof of diuen ansta statement let the polynomial be о лани юм. Илл Au pot Now will we will apply induction on n. in so nzo then there is no sign change so yo no coefficients are alternately positive negative which is true .. y nad Then there is a sign change and polynomial will look like pia - po ipo, po to & coefficients are alternately positive and negative Now, we the statement be true for nint degree polynomial and we will prove forn.. @ Scanned with rated GamScanner

roots then by undruction may be comprometice : polynomial of degree in having a positive be written as (x-2) 2 (-ign-1-j bj xd ae and p; are all positive nal 120 ; in left term we will put gal 1. un place of a =x n Ely ni ni nini •tu b -a 21 pax su po Dal 21 - n-n-2 h Po-" + ( bi ро-, "2 = Guello + E EN" dzi + a F-1 21 121 - wy"apo + (3, mas apent act* *ny JA +60% Pn-xh = (-v"copo + {1,*** [-»*° pou + ax kj] 2.° 3+ renten - tu epo + 3"{- ,so [ pain + opy I zi 3 Pinggin which also has non-zeus coefficients with alternating signs. rence, the given statement is true for n also.. del Tak siçanned with anos canner