Menu Close

Category: Algebra

Solve-for-x-R-3-x-x-3-x-

Question Number 161035 by cortano last updated on 11/Dec/21 $${Solve}\:{for}\:{x}\:\epsilon\:\mathbb{R}\: \\ $$$$\:\sqrt{\sqrt{\mathrm{3}}−{x}}\:=\:{x}\sqrt{\sqrt{\mathrm{3}}+{x}}\: \\ $$ Answered by Rasheed.Sindhi last updated on 11/Dec/21 $$\:\sqrt{\sqrt{\mathrm{3}}−{x}}\:=\:{x}\sqrt{\sqrt{\mathrm{3}}+{x}}\: \\ $$$$\sqrt{\mathrm{3}}\:−{x}={x}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}\:+{x}\right)…

1-x-1-x-x-3-0-

Question Number 161000 by bobhans last updated on 10/Dec/21 $$\:\:\sqrt{\mathrm{1}−\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}}}\:+\:\mid\mathrm{x}−\mathrm{3}\mid\:\geqslant\:\mathrm{0}\: \\ $$ Commented by cortano last updated on 10/Dec/21 $$\:\:\:\mathrm{x}<\mathrm{0} \\ $$ Terms of Service…

9b-2-25-why-is-this-inside-the-bracket-as-it-is-a-diffetence-of-two-squares-

Question Number 95469 by 6532 last updated on 25/May/20 $$\left(\mathrm{9b}^{\mathrm{2}} −\mathrm{25}\right) \\ $$$$\mathrm{why}\:\mathrm{is}\:\mathrm{this}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{bracket}\:\mathrm{as}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{diffetence}\:\mathrm{of}\:\mathrm{two}\:\mathrm{squares}? \\ $$ Commented by PRITHWISH SEN 2 last updated on 25/May/20 $$\left(\mathrm{3b}−\mathrm{5}\right)\left(\mathrm{3b}+\mathrm{5}\right)…

can-I-write-the-solution-of-ay-by-cy-0-y-c-1-e-b-b-2-4ac-2-x-c-2-e-b-b-2-4ac-2-x-when-b-2-4ac-0-c-1-e-b-2-x-c-2-xe-b-2-x-when-b-2-4ac-0-in-on

Question Number 95447 by Tony Lin last updated on 25/May/20 $${can}\:{I}\:{write}\:{the}\:{solution}\:{of} \\ $$$${ay}''+{by}'+{cy}=\mathrm{0} \\ $$$${y}=\begin{cases}{{c}_{\mathrm{1}} {e}^{\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}}{x}} +{c}_{\mathrm{2}} {e}^{\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}}{x}} ,{when}\:{b}^{\mathrm{2}} −\mathrm{4}{ac}\neq\mathrm{0}}\\{{c}_{\mathrm{1}} {e}^{\frac{−{b}}{\mathrm{2}}{x}} +{c}_{\mathrm{2}} {xe}^{\frac{−{b}}{\mathrm{2}}{x}}…

Question-160981

Question Number 160981 by mathlove last updated on 10/Dec/21 Commented by cortano last updated on 10/Dec/21 $$\:\mathrm{x}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)} \:=\:\sqrt{\mathrm{x}+\mathrm{1}}\: \\ $$$$\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)=\:\mathrm{log}\:_{\mathrm{3}} \left(\sqrt{\mathrm{x}+\mathrm{1}}\right) \\…