If-f-x-x-2-8-x-a-8-1-has-2-diffrent-real-roots-in-7a-find-the-range-of-a-gt-0-

Question Number 185764 by CrispyXYZ last updated on 27/Jan/23

If f(x)=−(x^2 /8)+x−(a/8)−1 has 2 diffrent real roots in ((√(7a)),+∞), find the range of a>0.

$$\mathrm{If}\:{f}\left({x}\right)=−\frac{{x}^{\mathrm{2}} }{\mathrm{8}}+{x}−\frac{{a}}{\mathrm{8}}−\mathrm{1}\:\mathrm{has}\:\mathrm{2}\:\mathrm{diffrent}\:\mathrm{real} \\ $$$$\mathrm{roots}\:\mathrm{in}\:\left(\sqrt{\mathrm{7}{a}},+\infty\right),\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{a}>\mathrm{0}. \\ $$

Answered by CrispyXYZ last updated on 28/Jan/23

Let g(x)=f(x+(√(7a))) = −(x^2 /8)+(1−((√(7a))/4))x−a+(√(7a))−1 (x>0, a>0) So we get: g(0)<0 and Δ>0 and symmetry axis=4−(√(7a))>0 ⇒a∈(0, ((5−(√(21)))/2))

$$\mathrm{Let}\:{g}\left({x}\right)={f}\left({x}+\sqrt{\mathrm{7}{a}}\right)\: \\ $$$$=\:−\frac{{x}^{\mathrm{2}} }{\mathrm{8}}+\left(\mathrm{1}−\frac{\sqrt{\mathrm{7}{a}}}{\mathrm{4}}\right){x}−{a}+\sqrt{\mathrm{7}{a}}−\mathrm{1}\:\left({x}>\mathrm{0},\:{a}>\mathrm{0}\right) \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{get}: \\ $$$${g}\left(\mathrm{0}\right)<\mathrm{0}\:\mathrm{and}\:\Delta>\mathrm{0}\:\mathrm{and}\:\mathrm{symmetry}\:\mathrm{axis}=\mathrm{4}−\sqrt{\mathrm{7}{a}}>\mathrm{0} \\ $$$$\Rightarrow{a}\in\left(\mathrm{0},\:\frac{\mathrm{5}−\sqrt{\mathrm{21}}}{\mathrm{2}}\right) \\ $$

Answered by Rajpurohith last updated on 27/Jan/23