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Question Number 27525 by sirigidiravikumar@gmail.com last updated on 08/Jan/18

x=7×4(√(3 ))thenx+1/x=?

$${x}=\mathrm{7}×\mathrm{4}\sqrt{\mathrm{3}\:}{thenx}+\mathrm{1}/{x}=? \\ $$

Commented by Rasheed.Sindhi last updated on 09/Jan/18

May be x=7+4(√(3 )) (Mistakenly written as x=7×4(√(3 ))) If x=7+4(√(3 )) x+(1/x)=(7+4(√(3 )) )+(1/(7+4(√(3 )) )) =(7+4(√(3 )) )+(1/(7+4(√(3 )) ))×((7−4(√(3 )) )/(7−4(√(3 )) )) =(7+4(√(3 )) )+((7−4(√(3 )) )/((7)^2 −(4(√(3 )) )^2 )) =7+4(√3)+((7−4(√(3 )) )/(49−48)) =7+4(√(3 )) +7−4(√(3 )) =14

$$\mathrm{May}\:\mathrm{be}\:{x}=\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}\:}\:\:\: \\ $$$$\left(\mathrm{Mistakenly}\:\mathrm{written}\:\mathrm{as}\:{x}=\mathrm{7}×\mathrm{4}\sqrt{\mathrm{3}\:}\right) \\ $$$$\mathrm{If}\:\mathrm{x}=\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}\:}\: \\ $$$$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}=\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}\:}\:\right)+\frac{\mathrm{1}}{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}\:}\:} \\ $$$$=\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}\:}\:\right)+\frac{\mathrm{1}}{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}\:}\:}×\frac{\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}\:}\:}{\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}\:}\:} \\ $$$$=\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}\:}\:\right)+\frac{\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}\:}\:}{\left(\mathrm{7}\right)^{\mathrm{2}} −\left(\mathrm{4}\sqrt{\mathrm{3}\:}\:\right)^{\mathrm{2}} } \\ $$$$=\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}+\frac{\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}\:}\:}{\mathrm{49}−\mathrm{48}} \\ $$$$=\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}\:}\:+\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}\:}\: \\ $$$$=\mathrm{14} \\ $$

Answered by Joel578 last updated on 09/Jan/18

x = 7 . 4(√3) = 28(√3) x + (1/x) = 28(√3) + (1/(28(√3))) = ((2353(√3))/(84)) ≈ 48.52

$${x}\:=\:\mathrm{7}\:.\:\mathrm{4}\sqrt{\mathrm{3}}\:=\:\mathrm{28}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{28}\sqrt{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{28}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{2353}\sqrt{\mathrm{3}}}{\mathrm{84}}\:\approx\:\mathrm{48}.\mathrm{52} \\ $$

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