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Category: Algebra

to-Sir-Aifour-we-can-construct-polynomes-of-both-3-rd-and-4-th-degree-in-a-way-that-the-constants-are-Z-or-Q-and-the-solutions-are-not-trivial-i-e-t-t-2-t-2-0-t-x-

Question Number 69479 by MJS last updated on 24/Sep/19 $$\mathrm{to}\:\mathrm{Sir}\:\mathrm{Aifour}: \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{construct}\:\mathrm{polynomes}\:\mathrm{of}\:\mathrm{both}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{and} \\ $$$$\mathrm{4}^{\mathrm{th}} \:\mathrm{degree}\:\mathrm{in}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{the}\:\mathrm{constants}\:\mathrm{are} \\ $$$$\in\mathbb{Z}\:\mathrm{or}\:\in\mathbb{Q}\:\mathrm{and}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{not}\:\mathrm{trivial} \\ $$$$\mathrm{i}.\mathrm{e}. \\ $$$$\left({t}−\alpha\right)\left({t}+\frac{\alpha}{\mathrm{2}}−\sqrt{\beta}\right)\left({t}+\frac{\alpha}{\mathrm{2}}+\sqrt{\beta}\right)=\mathrm{0}\wedge{t}={x}+\frac{\gamma}{\mathrm{3}} \\ $$$$\Leftrightarrow \\…

a-1-b-1-c-1-2abc-a-b-c-N-

Question Number 134981 by oooooooo last updated on 09/Mar/21 $$\left(\mathrm{a}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{b}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{c}}+\mathrm{1}\right)=\mathrm{2}\boldsymbol{\mathrm{abc}}\: \\ $$$$\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}\:\varepsilon\:\mathrm{N} \\ $$ Answered by MJS_new last updated on 10/Mar/21 $$\mathrm{let}\:{a}\leqslant{b}\leqslant{c} \\ $$$$\left(\mathrm{1}\right)\:{c}=\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right) \\…

Question-69411

Question Number 69411 by ahmadshah last updated on 23/Sep/19 Commented by Prithwish sen last updated on 23/Sep/19 $$\mathrm{h}+\mathrm{t}−\mathrm{c}=\mathrm{85}….\left(\mathrm{i}\right) \\ $$$$\mathrm{h}+\mathrm{t}+\mathrm{c}=\mathrm{155}…\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{ii}\right)−\left(\mathrm{i}\right)\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}}\:\:\boldsymbol{\mathrm{c}}=\:\mathrm{35}\:\boldsymbol{\mathrm{cm}}\:\: \\ $$ Terms…

Question-69413

Question Number 69413 by ahmadshah last updated on 23/Sep/19 Commented by kaivan.ahmadi last updated on 23/Sep/19 $$\frac{\left(\mathrm{5}×\frac{\mathrm{4}}{\mathrm{5}}\right)^{{log}_{\frac{\mathrm{5}}{\mathrm{4}}} \mathrm{4}} }{\mathrm{5}^{{log}_{\frac{\mathrm{5}}{\mathrm{4}}} \mathrm{5}} }=\frac{\mathrm{5}^{{log}_{\frac{\mathrm{5}}{\mathrm{4}}} \mathrm{4}} ×\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{log}_{\frac{\mathrm{5}}{\mathrm{4}}} \mathrm{4}} }{\mathrm{5}^{{log}_{\frac{\mathrm{5}}{\mathrm{4}}}…