Question Number 163285 by mnjuly1970 last updated on 05/Jan/22
$$ \\ $$$$\:\:\:\:#\:\mathrm{Q}{uestion}\:# \\ $$$$\:\:\:\:\:{suppose}\:{that}\:\:{x}_{\mathrm{1}} \:,\:\:{x}_{\:\mathrm{2}} \:\:{are}\:{two}\:{distinct} \\ $$$$\:\:\:\:{roots}\:{for}\:\:\:{ax}^{\:\mathrm{2}} +\:{bx}\:+{c}\:=\:\mathrm{0}\:\:{on}\:\left(\:\mathrm{0},\:\mathrm{1}\:\right). \\ $$$$\:\:\:\:\:{find}\:\:{the}\:{minimum}\:{value}\:{of}\:\:''\:{a}\:''\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}_{\:{min}} \:=\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−−−− \\ $$$$\:\:\:\:\:\: \\ $$
Answered by ajfour last updated on 05/Jan/22
$$\mathrm{2}{ax}_{\mathrm{1}} {x}_{\mathrm{2}} +{b}\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)+\mathrm{2}{c}=\mathrm{0} \\ $$$$\mathrm{2}{c}−\frac{{b}^{\mathrm{2}} }{{a}}+\mathrm{2}{c}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{a}=\frac{{b}^{\mathrm{2}} }{\mathrm{4}{c}} \\ $$