WHAT ABOUT THE SQUARE ROOT OF 20 AND A QUATER USING THE FACTORIZATION METHOD SORRY IM JUST REALLY BAD AT FACTORIZATION
Answers
Answer:
NOTE: √20 may be simplified to √[(4)(5)] ... Every positive real number has TWO square roots; One is positive and one is negative
Step-by-step explanation:
The prime factorisation is:
#20 = 2^2*5#
Hence:
#sqrt(20) = 2sqrt(5)#
This is an irrational number between #4# and #5# since:
#4^2 = 16 < 20 < 25 = 5^2#
It is not expressible as an exact fraction, but we can find rational approximations...
Since #20# is roughly halfway between #4^2# and #5^2#, its square root is approximately #9/2# - halfway between #4# and #5#.
In fact, we find:
#9^2 = 81 = 80+1 = 20*2^2 + 1#
which is in Pell's equation form:
#p^2 = n q^2 + 1#
with #n = 20#, #p = 9# and #q = 2#
That means that we can deduce the continued fraction for #sqrt(20)# from the continued fraction for #9/2#...
#9/2 = 4+1/2 = [4;2]#
Hence:
#sqrt(20) = [4;bar(2,8)] = 4+1/(2+1/(8+1/(2+1/(8+1/(2+1/(8+...))))))#
To get a good approximation for #sqrt(20)# truncate this continued fraction early, just before an '#8#'...
For example:
#sqrt(20) ~~ [4;2,8,2,8,2] = 4+1/(2+1/(8+1/(2+1/(8+1/2)))) = 2889/646 ~~ 4.472136#
From a calculator:
#sqrt(20) ~~ 4.472135955#