Theory of Equation- Descarte's rule of Sign In Hindi(Part 6)

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[@Wingsindigo883]

Sir meri ek query h...
X³-18x-35=0 ek equation hai...iske roots x=5 , x=(-5+i√3)/2 , x=(-5-i√3)/2 hain....one real root and two complex roots...
But jab hum Descarte's rule of sign apply kr rhe hein to yeh f(-x) ke liye two negative real roots bta raha h..jabki asliyt m iske pass 2 complex roots hein...plzz sir mere iss doubt ko solve kijiye

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See in playlist of theory of equations.

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It's really helpful to understand about Descartes rule of sign.

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Remove my doubts, sir. Q- x^2+x+1=0, why is no. of --ve roots=2 ?

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[@gauravjagtap5922]

positive roots will be at most 1 and not at least 1, and also for negative. Also there is one more relation that difference between degree and positive zeros is always even

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*Sir in question 1st why there atleast* ?? if there atleast four imag. then more will be because atleast four . if this then positive real root no. decrease but if it then it will not follow the rule that no. of change in sign equal to no. of positive real root? REPLY REPLY REPLY PLZZZZZZZZZ.......

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11.34 its not "atleast 1 real positive root" i think it will be " atmost 1 real positive root" please correct me if i am wrong

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If equation has p sign changes then it has p or p minus 2 real positive roots...how minus 2 comes ....plz sir explain

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Part 7 hai kiya is chennel mai???

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Sir agr koi positive roots kisi value ke between ho to kaise kre solve krenge

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[@syedmdabid7191]

Suppose that we have 2 +ve roots by this mthod in maximum. It means the + ve roots may be less than 2 or one root or no + ve root., because maximum +ve root is 2.This is result.

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sir. please upload the STRUMP THEORY...

[@mdasifhasan6531]

11:05 Sir in question no. 3rd, according to Descartes rule there may be two possibilities of roots. "4 complex roots" or " 1 positive root, 1 negative root, and 2 complex roots".