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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

$${consider}\:{f}\left({x}\right)={Ax}^{\mathrm{2}} +{Bx}+{C} \\ $$ $${with}\:{A}>\mathrm{0}.\:{Show}\:{that}\:{f}\left({x}\right)\geqslant\mathrm{0}\:\forall{x}\:\:{iff}\: \\ $$ $${B}^{\mathrm{2}} −\mathrm{4}{AC}\leqslant\mathrm{0} \\ $$

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

$${f}\left({x}\right)\geqslant\mathrm{0}\rightarrow{Ax}^{\mathrm{2}} +{Bx}+{C}\geqslant\mathrm{0} \\ $$ $${Ax}^{\mathrm{2}} +{Bx}+{C}={A}\left[\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} −\frac{{B}^{\mathrm{2}} }{\mathrm{4}{A}^{\mathrm{2}} }+\frac{{C}}{{A}}\right]\geqslant\mathrm{0} \\ $$ $${A}\left[\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} −\frac{{B}^{\mathrm{2}} }{\mathrm{4}{A}^{\mathrm{2}} }+\frac{\mathrm{4}{AC}}{\mathrm{4}{A}^{\mathrm{2}} }\right]\geqslant\mathrm{0} \\ $$ $${A}\left[\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} −\frac{{B}^{\mathrm{2}} −\mathrm{4}{AC}}{\mathrm{4}{A}^{\mathrm{2}} }\right]\geqslant\mathrm{0} \\ $$ $${then}\:\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} −\frac{{B}^{\mathrm{2}} −\mathrm{4}{AC}}{\mathrm{4}{A}^{\mathrm{2}} }\geqslant\mathrm{0}\:\:\:{because}\:{A}>\mathrm{0} \\ $$ $${now}\:{if}\:{B}^{\mathrm{2}} −\mathrm{4}{AC}\leqslant\mathrm{0}\rightarrow−\frac{{B}^{\mathrm{2}} −\mathrm{4}{AC}}{\mathrm{4}{A}^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$ $${and}\:\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$ $${we}\:{have}\:{A}\left[\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} −\frac{{B}^{\mathrm{2}} −\mathrm{4}{AC}}{\mathrm{4}{A}^{\mathrm{2}} }\right]\geqslant\mathrm{0} \\ $$ $${for}\:{B}^{\mathrm{2}} −\mathrm{4}{AC}\leqslant\mathrm{0}\:\:{with}\:{A}>\mathrm{0} \\ $$ $${finally}\:{f}\left({x}\right)\geqslant\mathrm{0} \\ $$ $$ \\ $$

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