How do you find the square root of 1728?

1 Answer
May 8, 2018

#sqrt(1728) = 24sqrt(3) ~~ 56451/1358 ~~ 41.5692194#

Explanation:

To find the square root of #1728#, first find its prime factorisation:

#color(white)(000)1728#
#color(white)(000)"/"color(white)(00)"\"#
#color(white)(00)2color(white)(000)864#
#color(white)(000000)"/"color(white)(0)"\"#
#color(white)(00000)2color(white)(00)432#
#color(white)(00000000)"/"color(white)(0)"\"#
#color(white)(0000000)2color(white)(00)216#
#color(white)(0000000000)"/"color(white)(0)"\"#
#color(white)(000000000)2color(white)(00)108#
#color(white)(000000000000)"/"color(white)(0)"\"#
#color(white)(00000000000)2color(white)(00)54#
#color(white)(0000000000000)"/"color(white)(00)"\"#
#color(white)(000000000000)2color(white)(000)27#
#color(white)(000000000000000)"/"color(white)(00)"\"#
#color(white)(00000000000000)3color(white)(0000)9#
#color(white)(000000000000000000)"/"color(white)(0)"\"#
#color(white)(00000000000000000)3color(white)(000)3#

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, #1728# actually has two square roots:

#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 #2^6# and #3^3# are both multiples of #3#, so #1728 = 2^6 * 3^3# has an exact cube root #2^2 * 3 = 12#

Approximations

If we want to approximate the value of #sqrt(1728) = 24sqrt(3)# with a good rational approximation, then we can proceed as follows...

Note that:

#7^2 = 49 = 48+1 = 3 * 4^2 + 1#

Hence a good first approximation for #sqrt(3)# is #7/4#.

Consider the quadratic with zeros #7+4sqrt(3)# and #7-4sqrt(3)#, namely:

#(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 #7+4sqrt(3)#.

Hence:

#sqrt(1728) = 6 * 4sqrt(3) ~~ 6*(37829/2716-7) = 6 * 18817/2716 = 56451/1358 ~~ 41.5692194#