Question Number 162366 by cortano last updated on 29/Dec/21
$$\:{Given}\:{that}\:{the}\:{solution}\:{set}\:{of}\:{the}\: \\ $$$$\:{quadratic}\:{inequality}\:{ax}^{\mathrm{2}} +{bx}+{c}\:>\mathrm{0} \\ $$$$\:{is}\:\left(\mathrm{2},\mathrm{3}\right).\:{Then}\:{the}\:{solution}\:{set}\: \\ $$$$\:{of}\:{the}\:{inequality}\:{cx}^{\mathrm{2}} +{bx}+{a}\:<\mathrm{0}\: \\ $$$$\:{will}\:{be}\: \\ $$
Answered by mr W last updated on 29/Dec/21
$${a}<\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\frac{{b}}{{a}}{x}+\frac{{c}}{{a}}=\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right) \\ $$$$\frac{{b}}{{a}}=−\left(\mathrm{2}+\mathrm{3}\right)=−\mathrm{5}\:\Rightarrow{b}=−\mathrm{5}{a} \\ $$$$\frac{{c}}{{a}}=\mathrm{2}×\mathrm{3}=\mathrm{6}\:\Rightarrow{c}=\mathrm{6}{a} \\ $$$${cx}^{\mathrm{2}} +{bx}+{a}={a}\left(\mathrm{6}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{1}\right)<\mathrm{0} \\ $$$${since}\:{a}<\mathrm{0}, \\ $$$$\Rightarrow\mathrm{6}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{1}>\mathrm{0} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{3}{x}−\mathrm{1}\right)>\mathrm{0} \\ $$$$\Rightarrow{x}<\frac{\mathrm{1}}{\mathrm{3}}\:{or}\:{x}>\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${i}.{e}.\:{x}\in\left(−\infty,\frac{\mathrm{1}}{\mathrm{3}}\right)\:\wedge\:\left(\frac{\mathrm{1}}{\mathrm{2}},+\infty\right) \\ $$