Question Number 20115 by Tinkutara last updated on 22/Aug/17 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{1}\:−\:{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} \:+\:\mathrm{2}{ax}\:−\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{has}\:\mathrm{roots} \\ $$$$\mathrm{belonging}\:\mathrm{to}\:\left(\mathrm{0},\:\mathrm{1}\right)\:\mathrm{is} \\ $$ Answered by ajfour last updated on 22/Aug/17…
Question Number 151181 by mathdanisur last updated on 18/Aug/21 $$\mathrm{if}\:\:\mid\boldsymbol{\mathrm{x}}\mid<\mathrm{1} \\ $$$$\mathrm{find}\:\:\mathrm{x}−\mathrm{4x}^{\mathrm{2}} +\mathrm{9x}^{\mathrm{3}} −\mathrm{16x}^{\mathrm{4}} +… \\ $$ Answered by Olaf_Thorendsen last updated on 18/Aug/21 $$\mathrm{S}\left({x}\right)\:=\:−\underset{{n}=\mathrm{1}}…
Question Number 151179 by mathdanisur last updated on 18/Aug/21 $$\mathrm{if}\:\:\mid\boldsymbol{\mathrm{x}}\mid<\mathrm{1} \\ $$$$\mathrm{find}\:\:\mathrm{x}+\mathrm{2x}^{\mathrm{2}} +\mathrm{3x}^{\mathrm{3}} +… \\ $$ Answered by Olaf_Thorendsen last updated on 18/Aug/21 $$\mathrm{S}\left({x}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 151174 by mathdanisur last updated on 18/Aug/21 $$\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 $$\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}×\mathrm{3}}}=\mathrm{3} \\ $$$${a}+\mathrm{3}=\mathrm{27}…
Question Number 85624 by Power last updated on 23/Mar/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151157 by kali last updated on 18/Aug/21 $${x} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151142 by mathdanisur last updated on 18/Aug/21 $$\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} \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{lies}\:\mathrm{in} \\ $$$$\left(\mathrm{1}\right)\:\left(−\mathrm{2},\:\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{1},\:\mathrm{1}\right) \\ $$$$\left(\mathrm{3}\right)\:\left(−\infty,\:−\mathrm{2}\right) \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{2},\:\infty\right) \\ $$…