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Question Number 189429 by 073 last updated on 16/Mar/23

Commented by 073 last updated on 16/Mar/23

solution please

$$\mathrm{solution}\:\mathrm{please} \\ $$

Answered by cortano12 last updated on 16/Mar/23

3∣x_1 ∣=∣x_2 ∣ ; x_1 ,x_2 <0 ⇒−3x_1 =−x_2 , x_2 =3x_1 ⇒x_1 +x_2 =−8 ⇒ { ((x_1 =−2)),((x_2 =−6)) :} ⇒f(x)=a(x+2)(x+6) ;(−4,8) ⇒8=a(−2)(2) ⇒a=−2 ∴ f(x)=−2(x^2 +8x+12) ⇒f(x)=−2x^2 −16x−24

$$\mathrm{3}\mid\mathrm{x}_{\mathrm{1}} \mid=\mid\mathrm{x}_{\mathrm{2}} \mid\:;\:\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} <\mathrm{0} \\ $$$$\Rightarrow−\mathrm{3x}_{\mathrm{1}} =−\mathrm{x}_{\mathrm{2}} \:,\:\mathrm{x}_{\mathrm{2}} =\mathrm{3x}_{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =−\mathrm{8} \\ $$$$\Rightarrow\begin{cases}{\mathrm{x}_{\mathrm{1}} =−\mathrm{2}}\\{\mathrm{x}_{\mathrm{2}} =−\mathrm{6}}\end{cases} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{6}\right)\:;\left(−\mathrm{4},\mathrm{8}\right) \\ $$$$\Rightarrow\mathrm{8}=\mathrm{a}\left(−\mathrm{2}\right)\left(\mathrm{2}\right)\:\Rightarrow\mathrm{a}=−\mathrm{2} \\ $$$$\therefore\:{f}\left({x}\right)=−\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{12}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{16}{x}−\mathrm{24} \\ $$

Commented by 073 last updated on 16/Mar/23

nice solution thanks alot please another one

$$\mathrm{nice}\:\mathrm{solution} \\ $$$$\mathrm{thanks}\:\mathrm{alot} \\ $$$$\mathrm{please}\:\mathrm{another}\:\mathrm{one} \\ $$

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