Prove-that-f-x-ax-2-bx-c-has-no-real-roots-if-and-only-if-a-f-b-2a-gt-0-

Question Number 115876 by ZiYangLee last updated on 29/Sep/20

Prove that f (x) = {a x}^{2} + b x + c has

no real roots if and only if a \cdot [f (- \frac{b}{2 a})] > 0

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

∙ a > 0, f (x) h a s n o r e a l r o o t s i i f b^{2} - 4 a c < 0

\Leftrightarrow f (- \frac{b}{2 a}) = a {(- \frac{b}{2 a})}^{2} + b (- \frac{b}{2 a}) + c = \frac{b^{2}}{4 a} - \frac{b^{2}}{2 a} + c

= - \frac{b^{2}}{4 a} + c = - \frac{b^{2} - 4 a c}{4 a} > 0

∙ a < 0, f (x) h a s n o r e a l s o l u t i o n s i i f f (- \frac{b}{2 a}) < 0 a s p r e v i o s l y s e e n \dots