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Question Number 88997 by jagoll last updated on 14/Apr/20

Commented by john santu last updated on 14/Apr/20

4R+2R(√2) = 8(√2)  R(2+(√2)) = 4(√2)  R = ((4(√2))/(2+(√2))) × ((2−(√2))/(2−(√2)))   R= ((8(√2)−8)/(4−2)) = 4(√2) −4 dm

$$\mathrm{4}{R}+\mathrm{2}{R}\sqrt{\mathrm{2}}\:=\:\mathrm{8}\sqrt{\mathrm{2}} \\ $$$${R}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\:=\:\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${R}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:×\:\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\: \\ $$$${R}=\:\frac{\mathrm{8}\sqrt{\mathrm{2}}−\mathrm{8}}{\mathrm{4}−\mathrm{2}}\:=\:\mathrm{4}\sqrt{\mathrm{2}}\:−\mathrm{4}\:{dm} \\ $$

Commented by mr W last updated on 14/Apr/20

Commented by mr W last updated on 14/Apr/20

2R=(√2)(4−R)  ⇒R=((4(√2))/(2+(√2)))=4((√2)−1)

$$\mathrm{2}{R}=\sqrt{\mathrm{2}}\left(\mathrm{4}−{R}\right) \\ $$$$\Rightarrow{R}=\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}=\mathrm{4}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$

Commented by $@ty@m123 last updated on 14/Apr/20

Wonderful!!!

$${Wonderful}!!! \\ $$

Commented by jagoll last updated on 15/Apr/20

waw.......very afdoll

$${waw}.......{very}\:{afdoll} \\ $$

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