Question Number 141002 by EnterUsername last updated on 14/May/21
$$\mathrm{If}\:{a}>\mathrm{0}\:\mathrm{and}\:\mathrm{one}\:\mathrm{root}\:\mathrm{of}\:{a}\mathrm{x}^{\mathrm{2}} +{b}\mathrm{x}+\mathrm{c}=\mathrm{0}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:−\mathrm{2} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{2},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}<\mathrm{0} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}>\mathrm{0} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}=\mathrm{0} \\ $$$$\left(\mathrm{D}\right)\:{a}+{b}={c} \\ $$
Answered by TheSupreme last updated on 14/May/21
$${x}_{\mathrm{1}} =\frac{−{b}−\sqrt{\Delta}}{\mathrm{2}{a}}<−\mathrm{2} \\ $$$${x}_{\mathrm{2}} =\frac{−{b}+\sqrt{\Delta}}{\mathrm{2}{a}}>\mathrm{2} \\ $$$$−{b}−\sqrt{\Delta}<−\mathrm{4}{a} \\ $$$$−{b}+\sqrt{\Delta}>\mathrm{4}{a} \\ $$$$\sqrt{\Delta}>\mathrm{4}{a}−{b} \\ $$$$\sqrt{\Delta}>\mathrm{4}{a}+{b} \\ $$$$\sqrt{\Delta}>\mathrm{4}{a}+\mid{b}\mid>\mathrm{0}\:\forall{x} \\ $$$${b}^{\mathrm{2}} −\mathrm{4}{ac}>\mathrm{16}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{8}{a}\mid{b}\mid \\ $$$$\mathrm{16}{a}^{\mathrm{2}} +\mathrm{8}{a}\mid{b}\mid+\mathrm{4}{ac}<\mathrm{0} \\ $$$$\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}<\mathrm{0}\:\left({A}\right) \\ $$$$ \\ $$
Commented by EnterUsername last updated on 19/May/21
$$\mathrm{Thanks} \\ $$