Question Number 151174 by mathdanisur last updated on 18/Aug/21 $$\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}\centerdot\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}\centerdot\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}\centerdot…}}}}}}\:\:=\:\mathrm{3}\: \\ $$$$\mathrm{find}\:\:{a}=? \\ $$ Answered by mr W last updated on 18/Aug/21 $$\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}×\mathrm{3}}}=\mathrm{3} \\ $$$${a}+\mathrm{3}=\mathrm{27}…
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Question Number 151142 by mathdanisur last updated on 18/Aug/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}\in\mathbb{R}^{+} \:\:\mathrm{and}\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{2}} }\:=\:\frac{\mathrm{27}}{\mathrm{4}} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} }\:+\:\frac{\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}}…
Question Number 20054 by Tinkutara last updated on 21/Aug/17 $$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\left(\alpha\:<\:\beta\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0},\:\mathrm{where} \\ $$$${c}\:<\:\mathrm{0}\:<\:{b},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{0}\:<\:\alpha\:<\:\beta \\ $$$$\left(\mathrm{2}\right)\:\alpha\:<\:\mathrm{0}\:<\:\beta\:<\:\mid\alpha\mid \\ $$$$\left(\mathrm{3}\right)\:\alpha\:<\:\beta\:<\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\alpha\:<\:\mathrm{0}\:<\:\mid\alpha\mid\:<\:\beta \\ $$…
Question Number 20052 by Tinkutara last updated on 21/Aug/17 $$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{are}\:\mathrm{real}\:\mathrm{and}\:\mathrm{of}\:\mathrm{opposite} \\ $$$$\mathrm{sign}\:\mathrm{then}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\alpha\left({x}\:−\:\beta\right)^{\mathrm{2}} \:+\:\beta\left({x}\:−\:\alpha\right)^{\mathrm{2}} \:\mathrm{is}/\mathrm{are} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Positive} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Negative} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Real}\:\mathrm{and}\:\mathrm{opposite}\:\mathrm{sign}…
Question Number 20053 by Tinkutara last updated on 21/Aug/17 $$\mathrm{If}\:\left(\mathrm{4}{a}\:+\:{c}\right)^{\mathrm{2}} \:\leqslant\:\mathrm{4}{b}^{\mathrm{2}} \:\mathrm{then}\:\mathrm{one}\:\mathrm{root}\:\mathrm{of} \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{lies}\:\mathrm{in} \\ $$$$\left(\mathrm{1}\right)\:\left(−\mathrm{2},\:\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{1},\:\mathrm{1}\right) \\ $$$$\left(\mathrm{3}\right)\:\left(−\infty,\:−\mathrm{2}\right) \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{2},\:\infty\right) \\ $$…
Question Number 20051 by Tinkutara last updated on 22/Aug/17 $$\mathrm{If}\:{x}\:\in\:{R}\:\mathrm{then}\:\frac{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:{a}}{{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:+\:\mathrm{3}{a}}\:\mathrm{can}\:\mathrm{take}\:\mathrm{all} \\ $$$$\mathrm{real}\:\mathrm{values}\:\mathrm{if} \\ $$$$\left(\mathrm{1}\right)\:{a}\:\in\:\left(\mathrm{0},\:\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:{a}\:\in\:\left[\mathrm{0},\:\mathrm{1}\right] \\ $$$$\left(\mathrm{3}\right)\:{a}\:\in\:\left[−\mathrm{1},\:\mathrm{1}\right] \\ $$$$\left(\mathrm{4}\right)\:\mathrm{None}\:\mathrm{of}\:\mathrm{these} \\ $$ Answered…
Question Number 20047 by tawa tawa last updated on 20/Aug/17 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}: \\ $$$$\frac{\sqrt{\mathrm{x}\:+\:\mathrm{1}}}{\mathrm{x}}\:+\:\sqrt{\frac{\mathrm{x}}{\mathrm{x}\:+\:\mathrm{1}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\ $$ Answered by ajfour last updated on 21/Jun/18 $${first}\:{term}\:{is}\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}\:\:{or}\:\:\frac{\sqrt{{x}+\mathrm{1}}}{{x}}\:\:? \\ $$$${assuming}\:{it}\:{is}\:\sqrt{\frac{\mathrm{1}+{x}}{{x}}}\:;…
Question Number 151115 by mathdanisur last updated on 18/Aug/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{0}}\\{\mathrm{x}^{\mathrm{2}} \mathrm{y}\:+\:\mathrm{xy}\:+\:\mathrm{1}\:=\:\mathrm{0}}\end{cases} \\ $$ Answered by dumitrel last updated on 18/Aug/21…