Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 32534 by naka3546 last updated on 27/Mar/18

Answered by MJS last updated on 27/Mar/18

m+(√n)=k f(x)=k x^2 +x−k=0 x=−(1/2)±(1/2)(√(4k+1)) f((1/x))=((3k+1)/(2k^2 ))±((k+1)/(2k^2 ))(√(4k+1)) case 1: +/+ case 2: +/− case 3: −/− case 1 ((3k+1)/k^2 )+((k+1)/k^2 )(√(4k+1))=(1/(10)) (k+1)(√(4k+1))=(1/(10))k^2 −3k−1 k^4 −460k^3 −20k^2 =0 of the 4 solutions only one is valid k=230+42(√(30)) ⇒ m=230; n=52920 100m+n=75920 case 2 ((3k+1)/k^2 )=(1/(10)) k^2 −30k−10=0 k=15−(√(235)) not valid ((√n)=−(√(235))) k=15+(√(235)) m=15; n=235 100m+n=1735 case 3 ((3k+1)/k^2 )+((k+1)/k^2 )(√(4k+1))=−(1/(10)) (k+1)(√(4k+1))=−(1/(10))k^2 −3k−1 no solution

$${m}+\sqrt{{n}}={k} \\ $$$${f}\left({x}\right)={k} \\ $$$${x}^{\mathrm{2}} +{x}−{k}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}{k}+\mathrm{1}} \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{3}{k}+\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} }\pm\frac{{k}+\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} }\sqrt{\mathrm{4}{k}+\mathrm{1}} \\ $$$$\mathrm{case}\:\mathrm{1}:\:+/+ \\ $$$$\mathrm{case}\:\mathrm{2}:\:+/− \\ $$$$\mathrm{case}\:\mathrm{3}:\:−/− \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{1} \\ $$$$\frac{\mathrm{3}{k}+\mathrm{1}}{{k}^{\mathrm{2}} }+\frac{{k}+\mathrm{1}}{{k}^{\mathrm{2}} }\sqrt{\mathrm{4}{k}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\left({k}+\mathrm{1}\right)\sqrt{\mathrm{4}{k}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{10}}{k}^{\mathrm{2}} −\mathrm{3}{k}−\mathrm{1} \\ $$$${k}^{\mathrm{4}} −\mathrm{460}{k}^{\mathrm{3}} −\mathrm{20}{k}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{4}\:\mathrm{solutions}\:\mathrm{only}\:\mathrm{one}\:\mathrm{is}\:\mathrm{valid} \\ $$$${k}=\mathrm{230}+\mathrm{42}\sqrt{\mathrm{30}} \\ $$$$\Rightarrow\:{m}=\mathrm{230};\:{n}=\mathrm{52920} \\ $$$$\mathrm{100}{m}+{n}=\mathrm{75920} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2} \\ $$$$\frac{\mathrm{3}{k}+\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{10}} \\ $$$${k}^{\mathrm{2}} −\mathrm{30}{k}−\mathrm{10}=\mathrm{0} \\ $$$${k}=\mathrm{15}−\sqrt{\mathrm{235}}\:\mathrm{not}\:\mathrm{valid}\:\left(\sqrt{{n}}=−\sqrt{\mathrm{235}}\right) \\ $$$${k}=\mathrm{15}+\sqrt{\mathrm{235}} \\ $$$${m}=\mathrm{15};\:{n}=\mathrm{235} \\ $$$$\mathrm{100}{m}+{n}=\mathrm{1735} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{3} \\ $$$$\frac{\mathrm{3}{k}+\mathrm{1}}{{k}^{\mathrm{2}} }+\frac{{k}+\mathrm{1}}{{k}^{\mathrm{2}} }\sqrt{\mathrm{4}{k}+\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\left({k}+\mathrm{1}\right)\sqrt{\mathrm{4}{k}+\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{10}}{k}^{\mathrm{2}} −\mathrm{3}{k}−\mathrm{1} \\ $$$$\mathrm{no}\:\mathrm{solution} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com