Prove that f(x)=ax^2 +bx+c has no real roots if and only if a∙[f(−(b/(2a)))]>0


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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

	$∙ a > 0, f (x) h a s n o r e a l r o o t s i i f b^{2} - 4 a c < 0$ $\Leftrightarrow f (- \frac{b}{2 a}) = a {(- \frac{b}{2 a})}^{2} + b (- \frac{b}{2 a}) + c = \frac{b^{2}}{4 a} - \frac{b^{2}}{2 a} + c$ $= - \frac{b^{2}}{4 a} + c = - \frac{b^{2} - 4 a c}{4 a} > 0$ $∙ a < 0, f (x) h a s n o r e a l s o l u t i o n s i i f f (- \frac{b}{2 a}) < 0 a s p r e v i o s l y s e e n . . .$
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