If a>0 and one root of ax^2 +bx+c=0 is less than −2 and the other is greater than 2, then (A) 4a+2∣b∣+c<0 (B) 4a+2∣b∣+c>0 (C) 4a+2∣b∣+c=0 (D) a+b=c


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Question Number 141002 by EnterUsername last updated on 14/May/21

$If a > 0 and one root of a x^{2} + b x + c = 0 is less than - 2$ $and the other is greater than 2, then$ $(A) 4 a + 2 ∣ b ∣ + c < 0$ $(B) 4 a + 2 ∣ b ∣ + c > 0$ $(C) 4 a + 2 ∣ b ∣ + c = 0$ $(D) a + b = c$
	Answered by TheSupreme last updated on 14/May/21

	$x_{1} = \frac{- b - \sqrt{Δ}}{2 a} < - 2$ $x_{2} = \frac{- b + \sqrt{Δ}}{2 a} > 2$ $- b - \sqrt{Δ} < - 4 a$ $- b + \sqrt{Δ} > 4 a$ $\sqrt{Δ} > 4 a - b$ $\sqrt{Δ} > 4 a + b$ $\sqrt{Δ} > 4 a + ∣ b ∣> 0 \forall x$ $b^{2} - 4 a c > 16 a^{2} + b^{2} + 8 a ∣ b ∣$ $16 a^{2} + 8 a ∣ b ∣ + 4 a c < 0$ $4 a + 2 ∣ b ∣ + c < 0 (A)$
	Commented by EnterUsername last updated on 19/May/21

	$Thanks$
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