What is the process to calculate #sqrt(0.9)#?

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Answer 1 · 2018-06-27T11:03:09

#sqrt(0.9) = sqrt (9/10)#

#sqrt (9/10) = [sqrt9]/[sqrt10]#

#3/[sqrt10] = [3sqrt10]/10#

Answer 2 · 2018-06-27T11:28:05

See below

We can use the Maclaurin series for #sqrt(1-x)# to approximate #sqrt0.9#.

The Maclaurin series for #sqrt(1-x)# is

#sqrt(1-x)=1-1/2x-1/8x^2-1/16x^3-...#

Plugging in #x=0.1# gives

#sqrt(1-0.1)=sqrt0.9=1-1/2(0.9)-1/8(0.9)^2...=0.94868#