How do you simplify #sqrt42# rounded to the nearest tenths place?

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Answer 1 · 2016-11-19T09:59:51

#sqrt(42) ~~ 6.5#

If you ask a calculator to find #sqrt(42)# then it will give you an approximation like:

#sqrt(42) ~~ 6.4807406984#

So this is between #6.4# and #6.5#, but much closer to #6.5# than #6.4#.

So rounded to the nearest tenths place, #sqrt(42) ~~ 6.5#

#color(white)()#
How would you find this without a calculator?

Note that:

#6*6 = 36#

#6*7 = 42#

#7*7 = 49#

#42# is nearly half way between #6^2 = 36# and #7^2 = 49#

So we expect #sqrt(42)# to be about halfway between #6# and #7#.

Looking at the graph of #y = x^2# for #x# between #6# and #7# and drawing a chord we see:

graph{(y-x^2)(y - (36+13(x-6))) = 0 [5.8, 7.2, 32, 51]}

By the time we get to #x in [6, 7]# the parabola is pretty straight, so we expect this linear interpolation to be good.

In fact we find:

#6.5^2 = 42.25#

Using Newton's method, the error in the approximation will be about:

#0.25 / (2*6.5) = 0.25/13 ~~ 0.02#

This is much smaller than #0.1# so will not affect the tenths place.

Another way of looking at this is continued fractions.

We find that any number of the form #n(n+1)# has a square root given by:

#sqrt(n(n+1)) = [n;bar(2,2n)] = n + 1/(2+1/(2n+1/(2+1/(2n+1/(2+...)))))#

In our example #42 = 6(6+1)#, so

#sqrt(42) = 6+1/(2+1/(12+1/(2+1/(12+1/(2+...)))))#

Truncating this, we can find rational approximations, such as:

#sqrt(42) ~~ [6;2] = 6+1/2 = 6.5#

#sqrt(42) ~~ [6;2,12] = 6+1/(2+1/12) = 6+12/25 = 6.48#