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

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

Answered by CrispyXYZ last updated on 28/Jan/23

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

Answered by Rajpurohith last updated on 27/Jan/23

Commented bymr W last updated on 28/Jan/23

$$wrong!$$ $$0<a<\frac{5-\sqrt{21}}{2}$$

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