Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 112087 by bobhans last updated on 06/Sep/20

$$\mathrm{a}\sqrt{\mathrm{a}}-3=10\sqrt{\mathrm{a}}\to \sqrt{\mathrm{a}}+\sqrt{{\mathrm{a}}^{-1}}=?$$

Answered by bemath last updated on 06/Sep/20

Answered by 1549442205PVT last updated on 06/Sep/20

$$\mathrm{Put}\sqrt{\mathrm{a}}=\mathrm{x}\Rightarrow \mathrm{x}⩾0.\mathrm{Then},\mathrm{from}\mathrm{the}$$$$\mathrm{hypothesis}\mathrm{a}\sqrt{\mathrm{a}}-3=10\sqrt{\mathrm{a}}\mathrm{we}\mathrm{have}$$$${\mathrm{x}}^3-10\mathrm{a}-3=0$$$$\iff (\mathrm{x}+3)({\mathrm{x}}^2-3\mathrm{x}-1)=0.\mathrm{Since}\mathrm{x}⩾0,$$$$\mathrm{x}+3>0\Rightarrow {\mathrm{x}}^2-3\mathrm{x}-1=0$$$$\iff \mathrm{x}=\frac{3+\sqrt{13}}{2}\Rightarrow \frac{1}{\mathrm{x}}=\frac{2}{3+\sqrt{13}}=\frac{2(3-\sqrt{13})}{(3+\sqrt{13})(3-\sqrt{13)}}$$$$=\frac{2(3-\sqrt{13})}{9-13}=-\frac{3-\sqrt{13}}{2}.\mathrm{Therefore},$$$${\mathbf{x}}_1+\frac{1}{{\mathbf{x}}_1}=\frac{3+\sqrt{13}}{2}-\frac{3-\sqrt{13}}{2}=\sqrt{13}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com