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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

$$\mathrm{Prove}\:\mathrm{that}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:\mathrm{has}\: \\ $$ $$\mathrm{no}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:{a}\centerdot\left[{f}\left(−\frac{{b}}{\mathrm{2}{a}}\right)\right]>\mathrm{0} \\ $$

Answered by Henri Boucatchou last updated on 29/Sep/20

   • a>0,  f(x)  has  no  real  roots  iif   b^2 −4ac<0      ⇔  f(−(b/(2a)))=a(−(b/(2a)))^2 +b(−(b/(2a)))+c=(b^2 /(4a))−(b^2 /(2a))+c  =−(b^2 /(4a))+c=−((b^2 −4ac)/(4a))>0     •  a<0,  f(x)  has  no  real  solutions  iif  f(−(b/(2a)))<0  as  previosly  seen...

$$\:\:\:\bullet\:{a}>\mathrm{0},\:\:{f}\left({x}\right)\:\:{has}\:\:{no}\:\:{real}\:\:{roots}\:\:{iif}\:\:\:{b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0} \\ $$ $$\:\:\:\:\Leftrightarrow\:\:{f}\left(−\frac{{b}}{\mathrm{2}{a}}\right)={a}\left(−\frac{{b}}{\mathrm{2}{a}}\right)^{\mathrm{2}} +{b}\left(−\frac{{b}}{\mathrm{2}{a}}\right)+{c}=\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}}−\frac{{b}^{\mathrm{2}} }{\mathrm{2}{a}}+{c} \\ $$ $$=−\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}}+{c}=−\frac{{b}^{\mathrm{2}} −\mathrm{4}{ac}}{\mathrm{4}{a}}>\mathrm{0} \\ $$ $$\:\:\:\bullet\:\:{a}<\mathrm{0},\:\:{f}\left({x}\right)\:\:{has}\:\:{no}\:\:{real}\:\:{solutions}\:\:{iif}\:\:{f}\left(−\frac{{b}}{\mathrm{2}{a}}\right)<\mathrm{0}\:\:{as}\:\:{previosly}\:\:{seen}... \\ $$

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