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

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

$$\mathrm{If}\:{a}>\mathrm{0}\:\mathrm{and}\:\mathrm{one}\:\mathrm{root}\:\mathrm{of}\:{a}\mathrm{x}^{\mathrm{2}} +{b}\mathrm{x}+\mathrm{c}=\mathrm{0}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:−\mathrm{2} \\ $$ $$\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{2},\:\mathrm{then} \\ $$ $$\left(\mathrm{A}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}<\mathrm{0} \\ $$ $$\left(\mathrm{B}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}>\mathrm{0} \\ $$ $$\left(\mathrm{C}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}=\mathrm{0} \\ $$ $$\left(\mathrm{D}\right)\:{a}+{b}={c} \\ $$

Answered by TheSupreme last updated on 14/May/21

x_1 =((−b−(√Δ))/(2a))<−2  x_2 =((−b+(√Δ))/(2a))>2  −b−(√Δ)<−4a  −b+(√Δ)>4a  (√Δ)>4a−b  (√Δ)>4a+b  (√Δ)>4a+∣b∣>0 ∀x  b^2 −4ac>16a^2 +b^2 +8a∣b∣  16a^2 +8a∣b∣+4ac<0  4a+2∣b∣+c<0 (A)

$${x}_{\mathrm{1}} =\frac{−{b}−\sqrt{\Delta}}{\mathrm{2}{a}}<−\mathrm{2} \\ $$ $${x}_{\mathrm{2}} =\frac{−{b}+\sqrt{\Delta}}{\mathrm{2}{a}}>\mathrm{2} \\ $$ $$−{b}−\sqrt{\Delta}<−\mathrm{4}{a} \\ $$ $$−{b}+\sqrt{\Delta}>\mathrm{4}{a} \\ $$ $$\sqrt{\Delta}>\mathrm{4}{a}−{b} \\ $$ $$\sqrt{\Delta}>\mathrm{4}{a}+{b} \\ $$ $$\sqrt{\Delta}>\mathrm{4}{a}+\mid{b}\mid>\mathrm{0}\:\forall{x} \\ $$ $${b}^{\mathrm{2}} −\mathrm{4}{ac}>\mathrm{16}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{8}{a}\mid{b}\mid \\ $$ $$\mathrm{16}{a}^{\mathrm{2}} +\mathrm{8}{a}\mid{b}\mid+\mathrm{4}{ac}<\mathrm{0} \\ $$ $$\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}<\mathrm{0}\:\left({A}\right) \\ $$ $$ \\ $$

Commented byEnterUsername last updated on 19/May/21

Thanks

$$\mathrm{Thanks} \\ $$

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