Question Number 172030 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$$\sqrt{\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6}}\:}\:\:+\:\:\mathrm{2}\:\sqrt{\frac{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}}\:=? \\ $$ Answered by Rasheed.Sindhi last updated on 23/Jun/22…
Question Number 172031 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$$\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}\:\:+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}\:=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}} \\ $$ Answered by puissant last updated on 23/Jun/22 $$\Rightarrow\left({x}−\mathrm{2}\right)+\left({x}+\mathrm{2}\right)=\sqrt{{x}^{\mathrm{2}}…
Question Number 172024 by Mikenice last updated on 23/Jun/22 $${let}\:\alpha\:{and}\:\beta\:{be}\:{the}\:{root}\:{of}\:{the}\:{equation} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}.\:{find}\:{the}\:{equation}\:{whose}\:{roots} \\ $$$${are}\:\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}\right)\:{and}\:\left(\frac{\mathrm{1}}{\alpha}−\frac{\mathrm{1}}{\beta}\right) \\ $$ Answered by Rasheed.Sindhi last updated on 23/Jun/22 $${Given}\:{equation}:…
Question Number 172025 by Mikenice last updated on 23/Jun/22 $${find}\:{the}\:{root}\:{of}\: \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{11}{x}^{\mathrm{2}} +\mathrm{14}{x}−\mathrm{30}=\mathrm{0}. \\ $$ Answered by thfchristopher last updated on 23/Jun/22 $${x}^{\mathrm{4}}…
Question Number 172027 by Mikenice last updated on 23/Jun/22 $${solve} \\ $$$$\frac{\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}=\frac{\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}} \\ $$ Commented by Mikenice last updated on 23/Jun/22 $${thanks}\:{sir}…
Question Number 172020 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$$\mathrm{2}^{{x}} =\mathrm{10}{x} \\ $$ Answered by mr W last updated on 23/Jun/22 $${e}^{{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{10}{x}…
Question Number 172021 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$$\mathrm{2}^{{x}^{\mathrm{2}} } ={x}^{\mathrm{2}{x}} \\ $$ Answered by puissant last updated on 23/Jun/22 $$\Rightarrow\:\:{x}^{\mathrm{2}} {ln}\mathrm{2}\:=\:\mathrm{2}{xlnx}…
Question Number 172022 by Mikenice last updated on 23/Jun/22 $${find}\:{the}\:{cubic}\:{of} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}}\:\:\:+\:\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}} \\ $$ Answered by Rasheed.Sindhi last updated on 23/Jun/22 $${Let}\:\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}}\:\:\:+\:\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}}\:={a} \\ $$$$\:{where}\:{a}\in\mathbb{R} \\…
Question Number 172023 by Mikenice last updated on 23/Jun/22 $${if}:\: \\ $$$${bx}^{\mathrm{3}} −\left(\mathrm{3}{b}+\mathrm{2}\right){x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{5}{b}−\mathrm{3}\right){x}+\mathrm{20}=\mathrm{0} \\ $$$${find}\:{b}\:{in}\:{term}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 172019 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$${x}^{\mathrm{3}} +\mathrm{9}{x}^{\mathrm{2}} {y}=\mathrm{10} \\ $$$${y}^{\mathrm{3}} +{xy}^{\mathrm{2}} =\mathrm{2} \\ $$ Terms of Service Privacy Policy…