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Question Number 145781 by Engr_Jidda last updated on 08/Jul/21
consider f(x)=Ax^2 +Bx+C with A>0. Show that f(x)≥0 ∀x iff B^2 −4AC≤0
considerf(x)=Ax2+Bx+CwithA>0.Showthatf(x)⩾0∀xiffB2−4AC⩽0
Answered by ArielVyny last updated on 08/Jul/21
f(x)≥0→Ax^2 +Bx+C≥0 Ax^2 +Bx+C=A[(x+(B/(2A)))^2 −(B^2 /(4A^2 ))+(C/A)]≥0 A[(x+(B/(2A)))^2 −(B^2 /(4A^2 ))+((4AC)/(4A^2 ))]≥0 A[(x+(B/(2A)))^2 −((B^2 −4AC)/(4A^2 ))]≥0 then (x+(B/(2A)))^2 −((B^2 −4AC)/(4A^2 ))≥0 because A>0 now if B^2 −4AC≤0→−((B^2 −4AC)/(4A^2 ))≥0 and (x+(B/(2A)))^2 ≥0 we have A[(x+(B/(2A)))^2 −((B^2 −4AC)/(4A^2 ))]≥0 for B^2 −4AC≤0 with A>0 finally f(x)≥0
f(x)⩾0→Ax2+Bx+C⩾0Ax2+Bx+C=A[(x+B2A)2−B24A2+CA]⩾0A[(x+B2A)2−B24A2+4AC4A2]⩾0A[(x+B2A)2−B2−4AC4A2]⩾0then(x+B2A)2−B2−4AC4A2⩾0becauseA>0nowifB2−4AC⩽0→−B2−4AC4A2⩾0and(x+B2A)2⩾0wehaveA[(x+B2A)2−B2−4AC4A2]⩾0forB2−4AC⩽0withA>0finallyf(x)⩾0

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