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 147587 by mathdanisur last updated on 22/Jul/21

Commented by ajfour last updated on 22/Jul/21

$$whatifwereplacether.h.s.in$$$$thisquestionby2x?$$

Answered by iloveisrael last updated on 22/Jul/21

$$\Rightarrow \mathrm{x}+\sqrt{\mathrm{kx}+2\mathrm{k}}+2\sqrt{{\mathrm{x}}^2-(\mathrm{kx}+2\mathrm{k})}+\mathrm{x}-\sqrt{\mathrm{kx}+2\mathrm{k}}={4\mathrm{k}}^2$$$$\Rightarrow 2\mathrm{x}-{4\mathrm{k}}^2=-2\sqrt{{\mathrm{x}}^2-\mathrm{kx}-2\mathrm{k}}$$$$\Rightarrow \mathrm{x}-{2\mathrm{k}}^2=-\sqrt{{\mathrm{x}}^2-\mathrm{kx}-2\mathrm{k}}$$$$\Rightarrow {\mathrm{x}}^2-{4\mathrm{k}}^2\mathrm{x}+{4\mathrm{k}}^4={\mathrm{x}}^2-\mathrm{kx}-2\mathrm{k}$$$$\Rightarrow ({4\mathrm{k}}^2-\mathrm{k})\mathrm{x}={4\mathrm{k}}^4+2\mathrm{k}$$$$\Rightarrow (4\mathrm{k}-1)\mathrm{x}={4\mathrm{k}}^3+2$$$$\Rightarrow \mathrm{x}=\frac{{4\mathrm{k}}^3+2}{4\mathrm{k}-1};\mathrm{k}\ne \frac{1}{4}$$

Commented by mathdanisur last updated on 22/Jul/21

$$ThankyouSir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com