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Question Number 63474 by aliesam last updated on 04/Jul/19

l e t P (x) = x^{2} + \frac{1}{2} x + b

a n d Q (x) = x^{2} + c x + d

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

P (x) Q (x) = Q (P (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 MJS last updated on 04/Jul/19

P (x) Q (x) - Q (P (x)) = 0

\Rightarrow

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

{\begin{cases} 4 c - 2 = 0 \\ 4 b + 2 c - 4 d + 1 = 0 \\ 4 b c - 4 b - 2 c + 2 d = 0 \\ 4 b^{2} + 4 b c - 4 b d + 4 d = 0 \end{cases}

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

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

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

P (Q (x)) = 0

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

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

\Rightarrow

x_{1} = - 1

x_{2} = \frac{1}{2}

x_{3, 4} = - \frac{1}{4} \pm \frac{\sqrt{15}}{4} i

Commented by aliesam last updated on 04/Jul/19

g o d b l e s s y o u

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