Menu Close

Category: Algebra

let-give-x-0-0-y-0-1-and-y-n-x-n-1-y-n-1-x-n-x-n-1-y-n-1-for-n-1-let-z-n-x-n-i-y-n-n-N-1-calculate-z-0-z-1-and-z-2-2-prove-that-n-N-n-1-z-n-1-i-z-n-

Question Number 31032 by abdo imad last updated on 02/Mar/18 $${let}\:{give}\:{x}_{\mathrm{0}} =\mathrm{0}\:,{y}_{\mathrm{0}} =\mathrm{1}\:{and}\:\left\{_{{y}_{{n}} ={x}_{{n}−\mathrm{1}} \:+{y}_{{n}−\mathrm{1}} } ^{{x}_{{n}} ={x}_{{n}−\mathrm{1}} \:−{y}_{{n}−\mathrm{1}} } \:\:\:\:\:\:{for}\:{n}\geqslant\mathrm{1}\:{let}\right. \\ $$$${z}_{{n}} ={x}_{{n}} \:+{i}\:{y}_{{n}}…

1-solve-inside-C-z-12-1-and-give-the-solution-at-form-r-e-i-2-calculate-1-u-u-2-u-n-then-find-the-solution-of-z-C-z-8-z-4-1-0-

Question Number 31031 by abdo imad last updated on 02/Mar/18 $$\left.\mathrm{1}\right)\:{solve}\:{inside}\:{C}\:\:{z}^{\mathrm{12}} =\mathrm{1}\:{and}\:{give}\:{the}\:{solution}\:{at}\:{form} \\ $$$${r}\:{e}^{{i}\theta} \\ $$$$\left.\mathrm{2}\right){calculate}\:\mathrm{1}+{u}\:+{u}^{\mathrm{2}} \:+…\:+{u}^{{n}} \:{then}\:{find}\:{the}\:{solution} \\ $$$${of}\:{z}\in{C}\:\:\:\:\:\:\:{z}^{\mathrm{8}} \:+{z}^{\mathrm{4}} \:+\mathrm{1}=\mathrm{0} \\ $$ Terms…

Solve-for-real-numbers-5-x-4-1-x-25-x-16-1-x-2527-

Question Number 162102 by HongKing last updated on 26/Dec/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{4}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} \:+\:\mathrm{25}^{\boldsymbol{\mathrm{x}}} \:\centerdot\:\mathrm{16}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} \:=\:\mathrm{2527} \\ $$ Commented by mr W last updated on…

Question-31018

Question Number 31018 by jasno91 last updated on 02/Mar/18 Commented by rahul 19 last updated on 02/Mar/18 $${sum}\:{of}\:{roots}\:=−\:\frac{{coefficient}\:{of}\:{b}}{{coefficient}\:{of}\:{a}} \\ $$$$\:{let}\:\mathrm{3}{rd}\:{root}=\:\zeta \\ $$$$\Rightarrow\:\sqrt{\mathrm{3}}\:+\left(\:−\sqrt{\mathrm{3}}\:\right)+\zeta\:=\:\frac{−\mathrm{1}}{\mathrm{2}}. \\ $$$${hence}\:\zeta\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:. \\…