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Question Number 195391 by mathlove last updated on 01/Aug/23

f^2 (x)+2f(x)=x^2 −8x+15 f(x)=?

$${f}^{\mathrm{2}} \left({x}\right)+\mathrm{2}{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{15} \\ $$$${f}\left({x}\right)=? \\ $$

Answered by MM42 last updated on 01/Aug/23

(f+1)^2 =x^2 −8x+16=(x−4)^2 ⇒f=∣x−4∣−1 ✓

$$\left({f}+\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{16}=\left({x}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{f}=\mid{x}−\mathrm{4}\mid−\mathrm{1}\:\checkmark \\ $$

Answered by kapoorshah last updated on 01/Aug/23

let f(x) = ax + b (ax + b)^2 + 2(ax + b) = x^2 − 8x +15 a^2 x^2 + 2abx + b^2 + 2ax + 2b = x^2 − 8x +15 a^2 x^2 + (2ab + 2a)x + b^2 + 2b = x^2 − 8x +15 a^2 = 1 a = 1 a = −1 2b + 2 = −8 −2b − 2 = −8 b = −5 b = 3 ∴ f(x) = x − 5 ∴ f(x) = −x + 3

$${let}\:{f}\left({x}\right)\:=\:{ax}\:+\:{b} \\ $$$$\left({ax}\:+\:{b}\right)^{\mathrm{2}} \:+\:\mathrm{2}\left({ax}\:+\:{b}\right)\:=\:{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\mathrm{15} \\ $$$${a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\:\mathrm{2}{abx}\:+\:{b}^{\mathrm{2}} \:+\:\mathrm{2}{ax}\:+\:\mathrm{2}{b}\:=\:{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\mathrm{15} \\ $$$${a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\:\left(\mathrm{2}{ab}\:+\:\mathrm{2}{a}\right){x}\:+\:{b}^{\mathrm{2}} \:+\:\mathrm{2}{b}\:=\:{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\mathrm{15} \\ $$$$ \\ $$$${a}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$${a}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\:=\:−\mathrm{1} \\ $$$$\mathrm{2}{b}\:+\:\mathrm{2}\:=\:−\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{b}\:−\:\:\mathrm{2}\:=\:−\mathrm{8} \\ $$$${b}\:=\:−\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}\:=\:\mathrm{3} \\ $$$$\therefore\:{f}\left({x}\right)\:=\:{x}\:−\:\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:{f}\left({x}\right)\:=\:−{x}\:+\:\mathrm{3} \\ $$$$ \\ $$

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