Menu Close

Category: Algebra

solve-simultaneously-2-k-h-9-i-k-2-h-3-ii-

Question Number 41984 by Tawa1 last updated on 16/Aug/18 $$\mathrm{solve}\:\mathrm{simultaneously}:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\sqrt{\mathrm{k}}\:\:+\:\mathrm{h}\:=\:\mathrm{9}\:\:\:…….\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{k}\:+\:\mathrm{2}\sqrt{\mathrm{h}}\:\:=\:\mathrm{3}\:\:…….\:\left(\mathrm{ii}\right) \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18 $${x}+\mathrm{2}\sqrt{{y}}\:−\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{2}\sqrt{{x}}\:+{y}−\mathrm{3}=\mathrm{0} \\…

Question-173054

Question Number 173054 by mathlove last updated on 06/Jul/22 Commented by Shrinava last updated on 06/Jul/22 $$\mathrm{a}+\mathrm{1}=\mathrm{x}\:\:,\:\:\mathrm{b}+\mathrm{1}=\mathrm{y}\:\:,\:\:\mathrm{c}+\mathrm{1}=\mathrm{z} \\ $$$$\mathrm{then}\:\:\:\mathrm{xyz}=\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=−\mathrm{6}\:\:\:\left(\mathrm{i}\right) \\ $$$$\Rightarrow\:\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{y}+\mathrm{2}\right)\left(\mathrm{z}+\mathrm{2}\right)=−\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{4x}+\mathrm{4y}+\mathrm{4z}+\mathrm{2xy}+\mathrm{2yz}+\mathrm{2zx}=−\mathrm{12}…

solve-x-2-x-3-x-4-x-6-5x-2-

Question Number 173051 by mr W last updated on 05/Jul/22 $${solve} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{6}\right)=\mathrm{5}{x}^{\mathrm{2}} \\ $$ Answered by dragan91 last updated on 06/Jul/22 $$\left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}\right)\left({x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}\right)−\mathrm{5}{x}^{\mathrm{2}}…

Question-173042

Question Number 173042 by mathlove last updated on 05/Jul/22 Answered by Rasheed.Sindhi last updated on 05/Jul/22 $$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}=\mathrm{3} \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\…

Question-41957

Question Number 41957 by behi83417@gmail.com last updated on 15/Aug/18 Commented by MJS last updated on 16/Aug/18 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{see}\:\mathrm{any}\:\mathrm{solution}\:\mathrm{with}\:\mathrm{all}\:\mathrm{variables}\:\in\mathbb{R} \\ $$$$\mathrm{in}\:\mathbb{C}\:\mathrm{we}\:\mathrm{can}\:\mathrm{freely}\:\mathrm{choose}\:\mathrm{3}\:\mathrm{variables}\:\mathrm{and}\:\mathrm{solve} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{remaining},\:\mathrm{so}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{unique}\:\mathrm{solution} \\ $$ Answered by…