Question Number 145781 by Engr_Jidda last updated on 08/Jul/21
$${consider}\:{f}\left({x}\right)={Ax}^{\mathrm{2}} +{Bx}+{C} \\ $$$${with}\:{A}>\mathrm{0}.\:{Show}\:{that}\:{f}\left({x}\right)\geqslant\mathrm{0}\:\forall{x}\:\:{iff}\: \\ $$$${B}^{\mathrm{2}} −\mathrm{4}{AC}\leqslant\mathrm{0} \\ $$
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](https://www.tinkutara.com/question/Q145784.png)
$${f}\left({x}\right)\geqslant\mathrm{0}\rightarrow{Ax}^{\mathrm{2}} +{Bx}+{C}\geqslant\mathrm{0} \\ $$$${Ax}^{\mathrm{2}} +{Bx}+{C}={A}\left[\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} −\frac{{B}^{\mathrm{2}} }{\mathrm{4}{A}^{\mathrm{2}} }+\frac{{C}}{{A}}\right]\geqslant\mathrm{0} \\ $$$${A}\left[\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} −\frac{{B}^{\mathrm{2}} }{\mathrm{4}{A}^{\mathrm{2}} }+\frac{\mathrm{4}{AC}}{\mathrm{4}{A}^{\mathrm{2}} }\right]\geqslant\mathrm{0} \\ $$$${A}\left[\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} −\frac{{B}^{\mathrm{2}} −\mathrm{4}{AC}}{\mathrm{4}{A}^{\mathrm{2}} }\right]\geqslant\mathrm{0} \\ $$$${then}\:\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} −\frac{{B}^{\mathrm{2}} −\mathrm{4}{AC}}{\mathrm{4}{A}^{\mathrm{2}} }\geqslant\mathrm{0}\:\:\:{because}\:{A}>\mathrm{0} \\ $$$${now}\:{if}\:{B}^{\mathrm{2}} −\mathrm{4}{AC}\leqslant\mathrm{0}\rightarrow−\frac{{B}^{\mathrm{2}} −\mathrm{4}{AC}}{\mathrm{4}{A}^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$$${and}\:\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${we}\:{have}\:{A}\left[\left({x}+\frac{{B}}{\mathrm{2}{A}}\right)^{\mathrm{2}} −\frac{{B}^{\mathrm{2}} −\mathrm{4}{AC}}{\mathrm{4}{A}^{\mathrm{2}} }\right]\geqslant\mathrm{0} \\ $$$${for}\:{B}^{\mathrm{2}} −\mathrm{4}{AC}\leqslant\mathrm{0}\:\:{with}\:{A}>\mathrm{0} \\ $$$${finally}\:{f}\left({x}\right)\geqslant\mathrm{0} \\ $$$$ \\ $$