deeper polynomial (1 Viewer)

[OLDMAN]

Not really difficult (like a HSC Q7 polynomial question, all bark no bite), but adds depth to your polynomial knowledge.

Let P(x) be the quadratic ax^2+bx+c. Suppose that P(x)=x has unequal roots. Show that the roots are also roots of P(P(x)=x. Find a quadratic equation for the other roots of this equation. Hence solve, (x^2-3x+2)^2-3(x^2-3x+2)+2-x=0.

 

[underthesun]

Hmm.. I think there's something wrong with P(P(x)=x

 

[Newbie]

inverse function?

 

[Archman]

P(a)=a
P(P(a))=P(a)=a

P(x)-x=a(x-p)(x-q)
P(P(x))-x=a^3(x-p)(x-q)(x-r)(x-s)

pq = c/a, pqrs=(ac^2 + bc + c)/a^3, rs = (ac + b + 1)/a^2
p + q = (1-b)/a, p+q+r+s = -2a^2*b/a^3= -2b/a, r+s = -b-1/a

therefore the required quadratic is:
a^2x^2+a(b+1)x+ac+b+1 = 0

(x^2-3x+2)^2-3(x^2-3x+2)+2-x = (x^2-3x+2-x)(x^2-2x)
and yeah.. just solve rhs

 

[Harimau]

I feel so inferior everytime i look at someone solving all these hard questions....

 

[OLDMAN]

Nice work Archman.
_________________________________________________
Quote Harimau:
I feel so inferior everytime i look at someone solving all these hard questions....
_________________________________________________

Don't be discouraged, a question like this in the HSC (Q7 or 8) comes in bite-sized parts. Anyone who made the effort understanding the problem and having a good go at it would have reaped some benefit.

 

[freaking_out]

Originally posted by Harimau
I feel so inferior everytime i look at someone solving all these hard questions....

don't worry there'll always b people smarter than us. ....:mad1: