how to find square root using binomial theorem
Answers
Answered by
1
hey there !!
Thus, the expansion of (1−2x)12(1−2x)12:
=1−x−12x2−12x3+...=1−x−12x2−12x3+...
The suggested way, is to choose a value for xx so that (1−2x)(1−2x) has the form 2∗2∗'a perfect square'. This can be done by taking x=0.01x=0.01. Thus, (1−2x)=(1−2∗0.01)=0.98=2∗0.72(1−2x)=(1−2∗0.01)=0.98=2∗0.72
And
(1−2x)12=0.9812=0.72–√(1−2x)12=0.9812=0.72
Which is equal to the previously established expansion, so we can now go ahead and find 2–√2. The problem I am facing, is that there was no mention of how the value of x=0.01x=0.01 was arrived at.
hope it helps : )
Thus, the expansion of (1−2x)12(1−2x)12:
=1−x−12x2−12x3+...=1−x−12x2−12x3+...
The suggested way, is to choose a value for xx so that (1−2x)(1−2x) has the form 2∗2∗'a perfect square'. This can be done by taking x=0.01x=0.01. Thus, (1−2x)=(1−2∗0.01)=0.98=2∗0.72(1−2x)=(1−2∗0.01)=0.98=2∗0.72
And
(1−2x)12=0.9812=0.72–√(1−2x)12=0.9812=0.72
Which is equal to the previously established expansion, so we can now go ahead and find 2–√2. The problem I am facing, is that there was no mention of how the value of x=0.01x=0.01 was arrived at.
hope it helps : )
Similar questions