How do you find the square root 28?

2 Answers
Sep 11, 2015

Answer:

Unfortunately the only easy way to do this is by using a calculator to get
#color(white)("XXX")sqrt(28)=5.291503#
(this is the principal [non-negative] square root)

Explanation:

If you must approximate #r=sqrt(28)# by hand:

Note

  • #5^2 = 25 < 28color(white)("XXX")# so #color(white)("XXX")r> 5#
  • #6^2= 36 > 28color(white)("XXX"# so #color(white)("XXX")r < 6#

taking a point between #5# and #6#
- #(5.5) = 30.25 > 28color(white)("XXX")# so #color(white)("XXX")r < 5.5#

taking a point between #5# and #5.5#
- #(5.3)^2 = 28.09 > 28color(white)("XXX")# so #color(white)("XXX")r < 5.3 " (but not a lot less)"#

Of course by now, you are probably wishing you had a calculator anyway to calculate the squared estimates...

Sep 11, 2015

Answer:

#sqrt(28) = sqrt(4*7) = sqrt(4)sqrt(7) = 2sqrt(7)#

As to calculating an approximation...

Explanation:

To calculate approximations of square roots of integers - especially if I'm trying to do it in my head or on a piece of paper - I like to find a sequence of rational approximations, stop at some point and long divide...

To do this I use a form of Newton Raphson method.

A standard formulation for finding the square root of a number #n# would be to start with an approximation #a_0#, then repeatedly apply a formula to give you better approximations:

#color(white)(X)a_(i+1) = (a_i^2 + n) / (2a_i)#

This is ok, but I prefer to split the numerator and denominator like this:

#color(white)(X)a_i = p_i/q_i#

...to keep things in terms of integers rather than fractions for as long as possible:

Then our formulas for iterating become:

#color(white)(X)p_(i+1) = p_i^2 + n q_i^2#

#color(white)(X)q_(i+1) = 2 p_i q_i#

If iterating in this way happens to give a #p_(i+1)# and #q_(i+1)# that have a common factor, then I divide both of them by that factor before the next iteration.

Let's do this with #n = 28#, #p_0 = 5# and #q_0 = 1# - that is starting with an approximation of #5# for #sqrt(28)# (since #5^2 = 25#).

#color(white)(X)p_1 = p_0^2 + n q_0^2 = 5^2 + (28 * 1^2) = 25+28 = 53#

#color(white)(X)q_1 = 2 p_0 q_0 = 2 * 5 * 1 = 10#

If we stopped at this stage we would have #sqrt(28) ~~ 53/10 = 5.3#

#color(white)(X)p_2 = p_1^2 + n q_1^2 = 53^2 + (28 * 10^2) = 2809 + 2800 = 5609#

#color(white)(X)q_2 = 2 * p_1 * q_1 = 2 * 53 * 10 = 1060#

If we stopped here, we would have #sqrt(28) ~~ 5609/1060 ~~ 5.2915#

#color(white)(X)p_3 = p_2^2 + n q_2^2 = 5609^2 + (28*1060^2)#

#color(white)(XX)= 31460881+31460800 = 62921681#

#color(white)(X)q_3 = 2 * p_2 * q_2 = 2 * 5609 * 1060 = 11891080#

Giving #sqrt(28) ~~ 62921681 / 11891080 ~~ 5.29150262213#

...accurate to #11# decimal places.

As you can see, this method converges quite fast, roughly doubling the number of significant digits on each iteration, but the numbers you are multiplying double in length on each iteration too.