consider f(x)=Ax^2 +Bx+C with A>0. Show that f(x)≥0 ∀x iff B^2 −4AC≤0


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Question Number 145781 by Engr_Jidda last updated on 08/Jul/21

$c o n s i d e r f (x) = {A x}^{2} + B x + C$ $w i t h A > 0 . S h o w t h a t f (x) ⩾ 0 \forall x i f f$ $B^{2} - 4 A C ⩽ 0$
	Answered by ArielVyny last updated on 08/Jul/21

	$f (x) ⩾ 0 \to {A x}^{2} + B x + C ⩾ 0$ ${A x}^{2} + B x + C = A [{(x + \frac{B}{2 A})}^{2} - \frac{B^{2}}{4 A^{2}} + \frac{C}{A}] ⩾ 0$ $A [{(x + \frac{B}{2 A})}^{2} - \frac{B^{2}}{4 A^{2}} + \frac{4 A C}{4 A^{2}}] ⩾ 0$ $A [{(x + \frac{B}{2 A})}^{2} - \frac{B^{2} - 4 A C}{4 A^{2}}] ⩾ 0$ $t h e n {(x + \frac{B}{2 A})}^{2} - \frac{B^{2} - 4 A C}{4 A^{2}} ⩾ 0 b e c a u s e A > 0$ $n o w i f B^{2} - 4 A C ⩽ 0 \to - \frac{B^{2} - 4 A C}{4 A^{2}} ⩾ 0$ $a n d {(x + \frac{B}{2 A})}^{2} ⩾ 0$ $w e h a v e A [{(x + \frac{B}{2 A})}^{2} - \frac{B^{2} - 4 A C}{4 A^{2}}] ⩾ 0$ $f o r B^{2} - 4 A C ⩽ 0 w i t h A > 0$ $f i n a l l y f (x) ⩾ 0$
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