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 85589 by sahnaz last updated on 23/Mar/20 $$\int\frac{\mathrm{1}+\mathrm{4u}}{−\mathrm{4u}^{\mathrm{2}} +\mathrm{2u}+\mathrm{2}}\mathrm{du} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20052 by Tinkutara last updated on 21/Aug/17 $$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{are}\:\mathrm{real}\:\mathrm{and}\:\mathrm{of}\:\mathrm{opposite} \\ $$$$\mathrm{sign}\:\mathrm{then}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\alpha\left({x}\:−\:\beta\right)^{\mathrm{2}} \:+\:\beta\left({x}\:−\:\alpha\right)^{\mathrm{2}} \:\mathrm{is}/\mathrm{are} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Positive} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Negative} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Real}\:\mathrm{and}\:\mathrm{opposite}\:\mathrm{sign}…
Question Number 20051 by Tinkutara last updated on 22/Aug/17 $$\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…
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Question Number 85582 by jagoll last updated on 23/Mar/20 $$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{9}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)= \\ $$ Commented by som(math1967) last updated on 23/Mar/20 $$\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{9}}\right) \\ $$ Terms of Service…
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 $${first}\:{term}\:{is}\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}\:\:{or}\:\:\frac{\sqrt{{x}+\mathrm{1}}}{{x}}\:\:? \\ $$$${assuming}\:{it}\:{is}\:\sqrt{\frac{\mathrm{1}+{x}}{{x}}}\:;…
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Question Number 20046 by Tinkutara last updated on 20/Aug/17 $$\mathrm{Anyone}\:\mathrm{who}\:\mathrm{has}\:\mathrm{a}\:\mathrm{very}\:\mathrm{high}\:\mathrm{passion} \\ $$$$\mathrm{for}\:\mathrm{Mathematics}\:\mathrm{should}\:\mathrm{join}\:\mathrm{AoPS} \\ $$$$\mathrm{community}.\:\mathrm{It}\:\mathrm{has}\:\mathrm{all}\:\mathrm{types}\:\mathrm{of}\:\mathrm{groups}, \\ $$$$\mathrm{from}\:\mathrm{middle}\:\mathrm{school}\:\mathrm{math}\:\mathrm{to}\:\mathrm{olympiad} \\ $$$$\mathrm{to}\:\mathrm{college}\:\mathrm{math}.\:\mathrm{But}\:\mathrm{a}\:\mathrm{new}\:\mathrm{user}\:\mathrm{can} \\ $$$$\mathrm{only}\:\mathrm{post}\:\mathrm{3}\:\mathrm{topics}.\:\mathrm{This}\:\mathrm{restriction}\:\mathrm{will} \\ $$$$\mathrm{be}\:\mathrm{removed}\:\mathrm{after}\:\mathrm{2}\:\mathrm{weeks}.\:\mathrm{Thanks}\:\mathrm{for} \\ $$$$\mathrm{joining}.\:\mathrm{Sign}\:\mathrm{up}\:\mathrm{on}: \\…