Question Number 185764 by CrispyXYZ last updated on 27/Jan/23
$$\mathrm{If}\:{f}\left({x}\right)=−\frac{{x}^{\mathrm{2}} }{\mathrm{8}}+{x}−\frac{{a}}{\mathrm{8}}−\mathrm{1}\:\mathrm{has}\:\mathrm{2}\:\mathrm{diffrent}\:\mathrm{real} \\ $$ $$\mathrm{roots}\:\mathrm{in}\:\left(\sqrt{\mathrm{7}{a}},+\infty\right),\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{a}>\mathrm{0}. \\ $$
Answered by CrispyXYZ last updated on 28/Jan/23
$$\mathrm{Let}\:{g}\left({x}\right)={f}\left({x}+\sqrt{\mathrm{7}{a}}\right)\: \\ $$ $$=\:−\frac{{x}^{\mathrm{2}} }{\mathrm{8}}+\left(\mathrm{1}−\frac{\sqrt{\mathrm{7}{a}}}{\mathrm{4}}\right){x}−{a}+\sqrt{\mathrm{7}{a}}−\mathrm{1}\:\left({x}>\mathrm{0},\:{a}>\mathrm{0}\right) \\ $$ $$\mathrm{So}\:\mathrm{we}\:\mathrm{get}: \\ $$ $${g}\left(\mathrm{0}\right)<\mathrm{0}\:\mathrm{and}\:\Delta>\mathrm{0}\:\mathrm{and}\:\mathrm{symmetry}\:\mathrm{axis}=\mathrm{4}−\sqrt{\mathrm{7}{a}}>\mathrm{0} \\ $$ $$\Rightarrow{a}\in\left(\mathrm{0},\:\frac{\mathrm{5}−\sqrt{\mathrm{21}}}{\mathrm{2}}\right) \\ $$
Answered by Rajpurohith last updated on 27/Jan/23
Commented bymr W last updated on 28/Jan/23
$${wrong}! \\ $$ $$\mathrm{0}<{a}<\frac{\mathrm{5}−\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$