Question Number 21321 by Tinkutara last updated on 20/Sep/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{4}{x}^{\mathrm{99}} \:+\:\mathrm{5}{x}^{\mathrm{98}} \:+\:\mathrm{4}{x}^{\mathrm{97}} \:+\:\mathrm{5}{x}^{\mathrm{96}} \:+ \\ $$$$…..\:+\:\mathrm{4}{x}\:+\:\mathrm{5}\:=\:\mathrm{0}\:\mathrm{is} \\ $$ Answered by dioph last updated…
Question Number 21319 by Tinkutara last updated on 20/Sep/17 $$\mathrm{If}\:{x},\:{y},\:{z}\:\mathrm{are}\:\mathrm{three}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such} \\ $$$$\mathrm{that}\:{x}\:+\:{y}\:+\:{z}\:=\:\mathrm{4}\:\mathrm{and}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{6}, \\ $$$$\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{2}}{\mathrm{3}}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{0}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{3} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{2}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{3}…
Question Number 21316 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Let}\:{p}\:=\:\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:….\:+ \\ $$$$\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{6}} \right)^{\mathrm{2}} \:+\:\left({x}_{\mathrm{2}} \:−\:{x}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:\left({x}_{\mathrm{2}} \:−\:{x}_{\mathrm{4}}…
Question Number 21314 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Let}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{be}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{px}\:−\:\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\:=\:\mathrm{0}, \\ $$$${p}\:\in\:{R}.\:\mathrm{The}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:\mathrm{is} \\ $$ Answered by $@ty@m last updated on 21/Sep/17…
Question Number 21315 by Tinkutara last updated on 20/Sep/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\sqrt[{\mathrm{4}}]{\mathrm{97}\:−\:{x}}\:+\:\sqrt[{\mathrm{4}}]{{x}}\:=\:\mathrm{5} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 21313 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Let}\:{k}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{inequality}\:\sqrt{{x}\:−\:\mathrm{3}}\:+\:\sqrt{\mathrm{6}\:−\:{x}}\:\geqslant\:{k}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{solution}\:\mathrm{then}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:{k} \\ $$$$\mathrm{is} \\ $$ Commented by mrW1 last updated on 20/Sep/17…
Question Number 21311 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Let}\:{a}\:\mathrm{and}\:{b}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\mathrm{with}\:{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:=\:{a}\:−\:{b},\:\mathrm{and}\:{k}\:=\:{a}^{\mathrm{2}} \:+\:\mathrm{4}{b}^{\mathrm{2}} , \\ $$$$\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{k}\:<\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:{k}\:>\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{k}\:=\:\mathrm{1} \\…
Question Number 21308 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{complex}\:\mathrm{numbers}\:{z}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mid{z}\:−\:\mid{z}\:+\:\mathrm{1}\mid\mid\:=\:\mid{z}\:+\:\mid{z}\:−\:\mathrm{1}\mid\mid \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 21309 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Suppose}\:{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{with}\:\mathrm{complex} \\ $$$$\mathrm{coefficients}\:\mathrm{and}\:\mathrm{an}\:\mathrm{even}\:\mathrm{degree}.\:\mathrm{If}\:\mathrm{all} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{p}\:\mathrm{are}\:\mathrm{complex}\:\mathrm{non}-\mathrm{real} \\ $$$$\mathrm{numbers}\:\mathrm{with}\:\mathrm{modulus}\:\mathrm{1},\:\mathrm{prove}\:\mathrm{that} \\ $$$${p}\left(\mathrm{1}\right)\:\in\:{R}\:\mathrm{iff}\:{p}\left(−\mathrm{1}\right)\:\in\:{R}. \\ $$ Terms of Service Privacy Policy…
Question Number 21307 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{such} \\ $$$$\mathrm{that} \\ $$$$\left(\mathrm{i}\right)\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{1} \\ $$$$\left(\mathrm{ii}\right)\:{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:\neq\:\mathrm{0}…