If-a-gt-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-lt-0-B-4a-2-b-c-gt-0-C-4a-2-b-c-0-D-a-b-c-

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