How do you find the square root of 1728?
1 Answer
Explanation:
To find the square root of
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So:
#1728 = 2^6 * 3^3#
and:
#sqrt(1728) = sqrt(2^6 * 3^3) = sqrt((2^3 * 3)^2 * 3) = 24sqrt(3)#
Note that like any positive number,
#sqrt(1728) = 24sqrt(3)" "# and#" "-sqrt(1728) = -24sqrt(3)#
When we say "the" square root, we normally mean the principal positive one.
Note that the powers in
Approximations
If we want to approximate the value of
Note that:
#7^2 = 49 = 48+1 = 3 * 4^2 + 1#
Hence a good first approximation for
Consider the quadratic with zeros
#(x-7-4sqrt(3))(x-7+4sqrt(3)) = x^2-14x+1#
From this we can define an integer sequence recursively:
#{ (a_0 = 0), (a_1 = 1), (a_(n+2) = 14a_(n+1)-a_n) :}#
The first few terms are:
#0, 1, 14, 195, 2716, 37829,...#
The ratio between successive terms rapidly converges toward
Hence:
#sqrt(1728) = 6 * 4sqrt(3) ~~ 6*(37829/2716-7) = 6 * 18817/2716 = 56451/1358 ~~ 41.5692194#