Quadratic eqns


@rudrasingh0
divide by $x_{3}$.
make a equation in x+1/x

@ashishdeosingh
I didn't get that

dividing by $x_{3}$,
= $x_{3}+1/x_{3}+a(x_{2}+1/x_{2})+b(x+1/x)+1=0$
$x_{2}+1/x_{2}=(x+1/x)_{2}−2$
$x_{3}+1/x_{3}=(x+1/x)((x+1/x)_{2}−3)$
put $x+1/x=t$
$t_{3}−3t+a(t_{2}−2)+bt+1=0$
$t_{3}+at_{2}+(b−3)t+1−2a=0$x=1 is root that is t = 2 is root
x=1 is not root that is t=2 is not root

@rudrasingh0 what's the answer?

@anonymouz Its 5

@ashishdeosingh thank you

@rudrasingh0 see the cubic in t will have one root =2... Equate x+1/x=2 we ll get 2 roots as 1 since it becomes a perfect square... But t=2 is not a root for x=1 won't be a repeated root... Now for that cubic in t we have one factor as t2... The remaining quadratic factor of t may give u 2 different roots that lie ( infi,  2)U(2,infi). If this happens then for each root of t we ll get 2 distinct roots of x.. Hence 5 roots are possible...
That interval is the range of x+1/x by Amgm

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What if we apply Des Cartes rule of sign and check for P(x) ?
Then we get 6 interval for sign change.
Which means there are 6 roots and one of them is 1. There are other 5 distinct real roots .

@omkarjungade
no.

@omkarjungade
You cannot say what is th sign of a and b