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

Commented by 073 last updated on 16/Mar/23

$solution please$
	Answered by cortano12 last updated on 16/Mar/23
	$3∣x_1 ∣=∣x_2 ∣ ; x_1 ,x_2 <0 ⇒−3x_1 =−x_2 , x_2 =3x_1 ⇒x_1 +x_2 =−8 ⇒ { ((x_1 =−2)),((x_2 =−6)) :} ⇒f(x)=a(x+2)(x+6) ;(−4,8) ⇒8=a(−2)(2) ⇒a=−2 ∴ f(x)=−2(x^2 +8x+12) ⇒f(x)=−2x^2 −16x−24$
	$3 ∣ x_{1} ∣=∣ x_{2} ∣; x_{1}, x_{2} < 0$ $\Rightarrow - {3 x}_{1} = - x_{2}, x_{2} = {3 x}_{1}$ $\Rightarrow x_{1} + x_{2} = - 8$ $\Rightarrow {\begin{cases} x_{1} = - 2 \\ x_{2} = - 6 \end{cases}$ $\Rightarrow f (x) = a (x + 2) (x + 6); (- 4, 8)$ $\Rightarrow 8 = a (- 2) (2) \Rightarrow a = - 2$ $∴ f (x) = - 2 (x^{2} + 8 x + 12)$ $\Rightarrow f (x) = - 2 x^{2} - 16 x - 24$
	Commented by 073 last updated on 16/Mar/23

	$nice solution$ $thanks alot$ $please another one$
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