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Question Number 110675 by Aina Samuel Temidayo last updated on 30/Aug/20

Two polynomials P and Q satisfy

P (- 2 x + Q (x)) = Q (- 2 x + P (x)) .

Given that Q (x) = x^{2} - 4 and

P (x) = ax + b . Find 2 a + b .

Answered by bemath last updated on 30/Aug/20

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

\Rightarrow P (- 2 x + Q (x)) = {ax}^{2} - 2 ax + 4 a + b

(2) - 2 x + P (x) = a (x - 2) + b

\Rightarrow Q (- 2 x + P (x)) = {a (x - 2) + b}^{2} - 4

{ax}^{2} - 2 ax + 4 a + b = a^{2} (x^{2} - 4 x + 4) + 2 a (x - 2) + b^{2}

{ax}^{2} - 2 ax + 4 a + b = a^{2} x^{2} - {4 a}^{2} x + {4 a}^{2} + 2 ax - 4 a + b^{2}

= a^{2} x^{2} + (2 a - {4 a}^{2}) x + {4 a}^{2} - 4 a + b^{2}

{\begin{cases} a = a^{2}, a = 0 or a = 1 \\ - 2 a = 2 a - {4 a}^{2} \Rightarrow {4 a}^{2} - 4 a = 0 \end{cases}

\to 4 a + b = {4 a}^{2} - 4 a + b^{2}

for a = 0 \to b^{2} - b = 0, b = 0 or 1

for a = 1 \Rightarrow 4 + b = b^{2}; b^{2} - b - 4 = 0

b = \frac{1 \pm \sqrt{17}}{2}

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

Thanks but with the options I have

here, the answer should be an integer .

Commented by bemath last updated on 30/Aug/20

what reason the answer should be integer ? given di condition ?

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

I {don}^{'} t understand

Commented by floor(10²Eta[1]) last updated on 30/Aug/20

Q (x) = x^{2} - 4 so why Q (x) - 2 x = x^{2} - 2 x + 4 ?

you also did {a (x - 2) + b}^{2} - 4 wrong

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

So can you solve it ?

Commented by bemath last updated on 30/Aug/20

why wrong ? Q (x) = x^{2} - 4 then

Q (x) - 2 x = x^{2} - 4 - 2 x = x^{2} - 2 x - 4

clear

Commented by floor(10²Eta[1]) last updated on 30/Aug/20

that is not what you type on the resolution

Answered by Aziztisffola last updated on 30/Aug/20

P (- 4) = b = Q (b) = b^{2} - 4

\Rightarrow b^{2} - b - 4 = 0 \Rightarrow b = \frac{1 \overset{+}{-} \sqrt{17}}{2}

P (- 2 + (- 3)) = Q (- 2 + a + b)

= {(a + b - 2)}^{2} - 4 = - 5 a + 6

\Rightarrow a^{2} + 2 ab + b^{2} + a - 5 b = 0

solve for a such that b = \frac{1 \overset{+}{-} \sqrt{17}}{2} .