Let-P-x-x-2-x-2-b-and-Q-x-x-2-cx-d-be-two-polynomial-with-real-coefficients-such-that-P-x-Q-x-Q-P-x-for-all-real-x-Find-all-the-real-roots-of-P-Q-x-0-

Question Number 194953 by dimentri last updated on 20/Jul/23

L e t P (x) = x^{2} + \frac{x}{2} + b a n d

Q (x) = x^{2} + c x + d b e t w o

p o l y n o m i a l w i t h r e a l c o e f f i c i e n t s

s u c h t h a t P (x) Q (x) = Q (P (x))

f o r a l l r e a l x .

F i n d a l l t h e r e a l r o o t s o f

P (Q (x)) = 0

Answered by Rasheed.Sindhi last updated on 21/Jul/23

P (x) = x^{2} + \frac{x}{2} + b, Q (x) = x^{2} + c x + d

P (x) Q (x) = Q (P (x))

R e a l r o o t s o f P (Q (x)) = 0 ?

P (x) Q (x) = Q (P (x))

l h s :

P (x) Q (x) = (x^{2} + \frac{x}{2} + b) (x^{2} + c x + d)

= x^{4} + {c x}^{3} + {d x}^{2}

+ \frac{1}{2} x^{3} + \frac{c}{2} x^{2} + \frac{d}{2} x

+ {b x}^{2} + b c x + b d

= x^{4} + (c + \frac{1}{2}) x^{3} + (b + d + \frac{c}{2}) x^{2} + (b c + \frac{d}{2}) x + b d

r h s : Q (P (x))

Q (P (x)) = {(x^{2} + \frac{x}{2} + b)}^{2} + c (x^{2} + \frac{x}{2} + b) + d

= x^{4} + \frac{1}{4} x^{2} + b^{2} + x^{3} + b x + 2 {b x}^{2} + {c x}^{2} + \frac{c}{2} x + b c + d

= x^{4} + x^{3} + \frac{1}{4} x^{2} + 2 {b x}^{2} + {c x}^{2} + b x + \frac{c}{2} x + b^{2} + b c + d

= x^{4} + x^{3} + (\frac{1}{4} + 2 b + c) x^{2} + (b + \frac{c}{2}) x + b^{2} + b c + d

∙ c + \frac{1}{2} = 1

c = \frac{1}{2}

∙ b + d + \frac{c}{2} = \frac{1}{4} + 2 b + c

b + d + \frac{1}{4} = \frac{1}{4} + 2 b + \frac{1}{2}

d = \frac{2 b + 1}{2}

∙ b c + \frac{d}{2} = b + \frac{c}{2}

\frac{b}{2} + \frac{2 b + 1}{4} = b + \frac{1}{4}

2 b + 2 b + 1 = 4 b + 1 N o r e s u l t

∙ b d = b^{2} + b c + d

b (\frac{2 b + 1}{2}) = b^{2} + \frac{b}{2} + \frac{2 b + 1}{2}

2 b^{2} + b = 2 b^{2} + b + 2 b + 1

b = - \frac{1}{2} \Rightarrow d = \frac{2 b + 1}{2} \Rightarrow d = 0

P (x) = x^{2} + \frac{x}{2} + b = x^{2} + \frac{x}{2} - \frac{1}{2} = \frac{2 x^{2} + x - 1}{2}

Q (x) = x^{2} + c x + d = x^{2} + \frac{1}{2} x = \frac{2 x^{2} + x}{2}

P (Q (x)) = 0

\frac{2 {(\frac{2 x^{2} + x}{2})}^{2} + (\frac{2 x^{2} + x}{2}) - 1}{2} = 0

2 {(\frac{2 x^{2} + x}{2})}^{2} + (\frac{2 x^{2} + x}{2}) - 1 = 0

\frac{4 x^{4} + 4 x^{3} + x^{2}}{2} + \frac{2 x^{2} + x}{2} - 1 = 0

4 x^{4} + 4 x^{3} + 3 x^{2} + x - 2 = 0

4 x^{4} + 4 x^{3} + 3 x^{2} + 3 x - 2 x - 2 = 0

4 x^{3} (x + 1) + 3 x (x + 1) - 2 (x + 1) = 0

(x + 1) (4 x^{3} + 3 x - 2) = 0

(x + 1) (2 x - 1) (2 x^{2} + x + 2) = 0

{\begin{cases} x + 1 = 0 \Rightarrow x = - 1 \\ 2 x - 1 = 0 \Rightarrow x = \frac{1}{2} \\ 2 x^{2} + x + 2 = 0 (n o r e a l r o o t s) \end{cases}

x = - 1, \frac{1}{2}

Commented by dimentri last updated on 21/Jul/23

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