Question Number 115876 by ZiYangLee last updated on 29/Sep/20
![Prove that f(x)=ax^2 +bx+c has no real roots if and only if a∙[f(−(b/(2a)))]>0](https://www.tinkutara.com/question/Q115876.png)
$$\mathrm{Prove}\:\mathrm{that}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:\mathrm{has}\: \\ $$$$\mathrm{no}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:{a}\centerdot\left[{f}\left(−\frac{{b}}{\mathrm{2}{a}}\right)\right]>\mathrm{0} \\ $$
Answered by Henri Boucatchou last updated on 29/Sep/20
$$\:\:\:\bullet\:{a}>\mathrm{0},\:\:{f}\left({x}\right)\:\:{has}\:\:{no}\:\:{real}\:\:{roots}\:\:{iif}\:\:\:{b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0} \\ $$$$\:\:\:\:\Leftrightarrow\:\:{f}\left(−\frac{{b}}{\mathrm{2}{a}}\right)={a}\left(−\frac{{b}}{\mathrm{2}{a}}\right)^{\mathrm{2}} +{b}\left(−\frac{{b}}{\mathrm{2}{a}}\right)+{c}=\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}}−\frac{{b}^{\mathrm{2}} }{\mathrm{2}{a}}+{c} \\ $$$$=−\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}}+{c}=−\frac{{b}^{\mathrm{2}} −\mathrm{4}{ac}}{\mathrm{4}{a}}>\mathrm{0} \\ $$$$\:\:\:\bullet\:\:{a}<\mathrm{0},\:\:{f}\left({x}\right)\:\:{has}\:\:{no}\:\:{real}\:\:{solutions}\:\:{iif}\:\:{f}\left(−\frac{{b}}{\mathrm{2}{a}}\right)<\mathrm{0}\:\:{as}\:\:{previosly}\:\:{seen}… \\ $$