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Question Number 20047 by tawa tawa last updated on 20/Aug/17

Solve for x: ((√(x + 1))/x) + (√(x/(x + 1))) = ((13)/6)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}: \\ $$$$\frac{\sqrt{\mathrm{x}\:+\:\mathrm{1}}}{\mathrm{x}}\:+\:\sqrt{\frac{\mathrm{x}}{\mathrm{x}\:+\:\mathrm{1}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\ $$

Answered by ajfour last updated on 21/Jun/18

first term is (√((x+1)/x)) or ((√(x+1))/x) ? assuming it is (√((1+x)/x)) ; let (√((1+x)/x)) =t then t+(1/t) = ((13)/6) 6t^2 −13t+6 = 0 6t^2 −4t−9t+6=0 2t(3t−2)−3(3t−2)=0 t= (3/2), (2/3) . ((x+1)/x) = (9/4), (4/9) 4x+4=9x or 9x+9=4x x= (4/5) or x=−(9/5) .

$${first}\:{term}\:{is}\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}\:\:{or}\:\:\frac{\sqrt{{x}+\mathrm{1}}}{{x}}\:\:? \\ $$$${assuming}\:{it}\:{is}\:\sqrt{\frac{\mathrm{1}+{x}}{{x}}}\:; \\ $$$${let}\:\sqrt{\frac{\mathrm{1}+{x}}{{x}}}\:={t} \\ $$$${then}\:\:\:\:\:{t}+\frac{\mathrm{1}}{{t}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\:\:\:\:\mathrm{6}{t}^{\mathrm{2}} −\mathrm{13}{t}+\mathrm{6}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{6}{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{9}{t}+\mathrm{6}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{2}{t}\left(\mathrm{3}{t}−\mathrm{2}\right)−\mathrm{3}\left(\mathrm{3}{t}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:{t}=\:\frac{\mathrm{3}}{\mathrm{2}},\:\frac{\mathrm{2}}{\mathrm{3}}\:. \\ $$$$\:\:\:\frac{{x}+\mathrm{1}}{{x}}\:=\:\frac{\mathrm{9}}{\mathrm{4}},\:\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\:\:\mathrm{4}{x}+\mathrm{4}=\mathrm{9}{x}\:\:\:\:{or}\:\:\:\:\mathrm{9}{x}+\mathrm{9}=\mathrm{4}{x} \\ $$$$\:\:\:\:\:\:\:\:{x}=\:\frac{\mathrm{4}}{\mathrm{5}}\:\:\:\:\:\:{or}\:\:\:\:\:\:\:\:\:{x}=−\frac{\mathrm{9}}{\mathrm{5}}\:. \\ $$

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