Question Number 175943 by CrispyXYZ last updated on 10/Sep/22
$$\left(\mathrm{1}\right)\:\exists{x}\in\mathbb{R},\:{x}^{\mathrm{2}} +{x}+{a}=\mathrm{0}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{a}. \\ $$$$\left(\mathrm{2}\right)\:\forall{x}\in\mathbb{R},\:{x}^{\mathrm{2}} +{x}+{a}=\mathrm{0}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{a}. \\ $$
Answered by mahdipoor last updated on 10/Sep/22
$$\left.\mathrm{1}\right)\:{x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}{a}}}{\mathrm{2}}\:\Rightarrow\:\mathrm{1}−\mathrm{4}{a}\geqslant\mathrm{0}\:\Rightarrow\:{a}\leqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left.\mathrm{2}\right){for}\:{example}\::\:{get}\:{x}_{\mathrm{1}} =\mathrm{1}\:\:{and}\:{x}_{\mathrm{0}} =\mathrm{0} \\ $$$$\left({x}_{\mathrm{1}} \right)^{\mathrm{2}} +{x}_{\mathrm{1}} +{a}=\left({x}_{\mathrm{0}} \right)^{\mathrm{2}} +{x}_{\mathrm{0}} +{a}=\mathrm{0} \\ $$$$\mathrm{2}+{a}={a}=\mathrm{0}\:\Rightarrow\:\nexists{a}\in{R} \\ $$