Question Number 20259 by Tinkutara last updated on 24/Aug/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$ Answered by ajfour last updated on 24/Aug/17 $${f}\:'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\:>\mathrm{0}…
Question Number 151322 by mathdanisur last updated on 20/Aug/21 $$\mathrm{Determine}\:\mathrm{all}\:\mathrm{pairs}\:\:\left(\mathrm{x};\mathrm{y}\right)\:\:\mathrm{of}\:\mathrm{integers} \\ $$$$\mathrm{which}\:\mathrm{satisfy}: \\ $$$$\mathrm{x}^{\mathrm{3}} \:-\:\mathrm{3x}\:+\:\mathrm{y}\:=\:\mathrm{0} \\ $$ Commented by mathdanisur last updated on 20/Aug/21 $$\mathrm{with}\:\boldsymbol{\mathrm{x}}\:\mathrm{real},\:\boldsymbol{\mathrm{y}}\:\mathrm{integer}…
Question Number 151316 by iloveisrael last updated on 20/Aug/21 $$\:\:\:\:\:\:{Find}\:{positive}\:{integers}\: \\ $$$$\:\:\:\:\:{a}\:{and}\:{b}\:{such}\:{that}\: \\ $$$$\:\:\:\:\:\left(\sqrt[{\mathrm{3}}]{{a}}\:+\sqrt[{\mathrm{3}}]{{b}}\:−\mathrm{1}\right)^{\mathrm{2}} =\:\mathrm{49}+\:\mathrm{20}\sqrt[{\mathrm{3}}]{\mathrm{6}}\: \\ $$ Answered by Rasheed.Sindhi last updated on 20/Aug/21 $$\overset{\mathcal{SOLVE}\:\:{for}\:{a},{b}\in\mathbb{Z}^{+}…
Question Number 151317 by mathdanisur last updated on 20/Aug/21 $$\mathrm{if}\:\:\boldsymbol{\mathrm{x}}=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}\:-\:\sqrt{\mathrm{6}}}\:+\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}\:+\:\sqrt{\mathrm{6}}} \\ $$$$\mathrm{find}\:\:\mathrm{E}\left(\mathrm{x}\right)\:=\:\mathrm{x}^{\mathrm{6}} \:-\:\mathrm{6x}^{\mathrm{4}} \:+\:\mathrm{9x}^{\mathrm{2}} \:=\:? \\ $$ Answered by iloveisrael last updated on 20/Aug/21 $$\:{we}\:{have}\:{x}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}−\sqrt{\mathrm{6}}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}+\sqrt{\mathrm{6}}}\:=\:\mathrm{0}…
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Question Number 151315 by mathdanisur last updated on 20/Aug/21 $$\mathrm{if}\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{16}^{\mathrm{2}} \:\:;\:\:\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{24}^{\mathrm{2}} \\ $$$$\mathrm{z}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} =\mathrm{42}^{\mathrm{2}} \:\:\mathrm{and}\:\:\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} =\mathrm{38}^{\mathrm{2}} \\ $$$$\mathrm{find}\:\:\mathrm{max}\left[\left(\mathrm{x}+\mathrm{z}\right)\left(\mathrm{y}+\mathrm{t}\right)\right]=? \\…
Question Number 151308 by mathdanisur last updated on 19/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151307 by mathdanisur last updated on 19/Aug/21 $$\mathrm{if}\:\:\mathrm{x}\sqrt{\mathrm{x}}\:-\:\mathrm{26}\sqrt{\mathrm{x}}\:=\:\mathrm{5} \\ $$$$\mathrm{find}\:\:\mathrm{x}\:-\:\mathrm{5}\sqrt{\mathrm{x}}\:=\:? \\ $$ Answered by MJS_new last updated on 20/Aug/21 $${x}\sqrt{{x}}−\mathrm{26}\sqrt{{x}}−\mathrm{5}=\mathrm{0} \\ $$$$\left(\sqrt{{x}}+\mathrm{5}\right)\left({x}−\mathrm{5}\sqrt{{x}}−\mathrm{1}\right)=\mathrm{0} \\…
Question Number 85762 by Power last updated on 24/Mar/20 Commented by MJS last updated on 24/Mar/20 $$\approx\mathrm{1}.\mathrm{43312742672} \\ $$ Commented by I want to learn…
Question Number 151300 by mathdanisur last updated on 19/Aug/21 $$\mathrm{5}\:\centerdot\:\mathrm{6},\mathrm{02}\centerdot\mathrm{10}^{\mathrm{23}} \:=\:?\:\left(\mathrm{solution}\right) \\ $$$$\left.\mathrm{a}\left.\right)\mathrm{3},\mathrm{01}\centerdot\mathrm{10}^{\mathrm{24}} \:\:\:\mathrm{b}\right)\mathrm{3},\mathrm{01}\centerdot\mathrm{10}^{\mathrm{22}} \\ $$ Answered by puissant last updated on 19/Aug/21 $$\mathrm{5}×\mathrm{6},\mathrm{02}.\mathrm{10}^{\mathrm{23}} =\mathrm{3},\mathrm{1}.\mathrm{10}^{\mathrm{24}}…