How do you find a one-decimal place approximation for #root4 45#?

Question

1620 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-16T19:16:33

#root(4)45=2.6# by calculator

#root(4)45=2.6# by calculator

Answer 2 · 2015-10-16T20:40:12

Use rational approximations for #sqrt(3)# and #sqrt(5)# to find

#root(4)(45) ~~ 2.6#

#root(4)(45) = sqrt(sqrt(45)) = sqrt(sqrt(3^2*5)) = sqrt(3) sqrt(sqrt(5))#

#sqrt(3) ~~ 7/4#

#sqrt(sqrt(5)) ~~ sqrt(9/4) = 3/2#

So

#root(4)(45) ~~ 7/4 * 3/2 = 21 / 8 = 2.625 ~~ 2.6#