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Category: Algebra

we-have-for-quadratic-equations-x-b-b-2-4ac-2a-what-about-cubic-equation-is-there-any-rules-or-ways-to-solve-

Question Number 93963 by  M±th+et+s last updated on 16/May/20 $${we}\:{have}\:{for}\:{quadratic}\:{equations} \\ $$$${x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${what}\:{about}\:{cubic}\:{equation}\:{is}\:{there}\:{any} \\ $$$${rules}\:{or}\:{ways}\:{to}\:{solve}? \\ $$ Commented by Kunal12588 last updated on…

Question-159494

Question Number 159494 by HongKing last updated on 17/Nov/21 Answered by qaz last updated on 18/Nov/21 $$\left(\mathrm{1}+\mathrm{a}\right)\left(\mathrm{1}+\frac{\mathrm{b}}{\mathrm{a}}\right)\left(\mathrm{1}+\frac{\mathrm{c}}{\mathrm{b}}\right)\left(\mathrm{1}+\frac{\mathrm{81}}{\mathrm{c}}\right) \\ $$$$\geqslant\left(\mathrm{1}+\sqrt{\mathrm{a}}\centerdot\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\frac{\mathrm{c}}{\mathrm{b}}}\centerdot\sqrt{\frac{\mathrm{81}}{\mathrm{c}}}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{1}+\sqrt{\mathrm{b}}\right)^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\frac{\mathrm{81}}{\mathrm{b}}}\right)^{\mathrm{2}} \\ $$$$\geqslant\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{b}}\centerdot\sqrt[{\mathrm{4}}]{\frac{\mathrm{81}}{\mathrm{b}}}\right)^{\mathrm{4}}…

Find-1-x-1-x-2-dx-

Question Number 159477 by HongKing last updated on 17/Nov/21 $$\mathrm{Find}:\:\:\:\Omega\:=\int\:\frac{\mathrm{1}}{\left(\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$ Answered by MJS_new last updated on 17/Nov/21 $${t}=\mathrm{arctan}\:{x} \\ $$$$\Rightarrow \\ $$$$\int\mathrm{sin}^{\mathrm{2}}…

Question-159479

Question Number 159479 by quvonnn last updated on 17/Nov/21 Answered by Tokugami last updated on 17/Nov/21 $$\frac{\mathrm{2}}{\mathrm{4}×\mathrm{2}!}+\frac{\mathrm{2}}{\mathrm{5}×\mathrm{3}!}+\frac{\mathrm{2}}{\mathrm{6}×\mathrm{4}!}+\frac{\mathrm{2}}{\mathrm{7}×\mathrm{5}!}+\frac{\mathrm{2}}{\mathrm{8}×\mathrm{6}!}+… \\ $$$$=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}+\mathrm{2}\right){n}!} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}\left({n}+\mathrm{1}\right)}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n}!}…

Question-159472

Question Number 159472 by mnjuly1970 last updated on 17/Nov/21 Answered by MJS_new last updated on 18/Nov/21 $$\mathrm{1}−{x}^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$${x}+\lfloor{x}\rfloor\geqslant\mathrm{0}\:\Rightarrow\:{x}\geqslant\mathrm{0} \\ $$$$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\sqrt{{x}+\lfloor{x}\rfloor}\geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{1}−{x}^{\mathrm{2}}…

Question-93916

Question Number 93916 by ajfour last updated on 16/May/20 Commented by mr W last updated on 16/May/20 $${it}'{s}\:{great}\:{sir}! \\ $$$${with}\:{b}=\mathrm{3},\:{c}=\mathrm{1}\:{i}\:{didn}'{t}\:{get}\:{the}\:{expected} \\ $$$${root}.\:{can}\:{you}\:{recheck}\:{please}? \\ $$ Commented…