How do you find the square root of 270?

3 Answers
Apr 29, 2017

Answer:

See the solution process below:

Explanation:

We can use this rule of radicals to rewrite this expression:

#sqrt(a * b) = sqrt(a) * sqrt(b)#

#sqrt(270) = sqrt(9 * 30) = sqrt(9) * sqrt(30) = +-3sqrt(30)#

If necessary, the #sqrt(30) = +-5.477#

And therefore:

#+-3sqrt(30) = +-3 *+- 5.477 = +-16.432# rounded to the nearest thousandth.

Apr 30, 2017

Answer:

A = #3 * sqrt30#

Explanation:

Find a #sqrt# which can be divided by the whole number:
= #sqrt30#

Then find the quotient of the number and the divisor (#sqrt#):
= 270 / #sqrt30#
= 3
= 3 * #sqrt30#

Apr 30, 2017

Answer:

#sqrt(270) = 3sqrt(30) ~~ 15873/966 ~~ 16.431677#

Explanation:

First note that #270#, like any positive number, has two square roots. We denote the positive square root by #sqrt(270)# and the negative one by #-sqrt(270)#. However, the expression "the square root" is often used to refer to the principal, positive square root.

Next note that if #a, b >= 0# then:

#sqrt(ab) = sqrt(a)sqrt(b)#

[[ The same is not true if both #a < 0# and #b < 0# ]]

Also, if #a >= 0# then:

#sqrt(a^2) = a#

So we find:

#sqrt(270) = sqrt(9*30) = sqrt(9)sqrt(30) = 3sqrt(30)#

This is the simplest form of the principal square root.

#3sqrt(30)#, like #sqrt(30)# is an irrational number.

We can calculate approximations to #sqrt(30)# using its continued fraction. Note that #30=5*6# is of the form #n(n+1)#. Hence its square root has a short regular repeating pattern:

#sqrt(30) = [5;bar(2,10)] = 5+1/(2+1/(10+1/(2+1/(10+1/(2+1/(10+...))))))#

[[ In general #sqrt(n(n+1)) = [n;bar(2,2n)]# ]]

We can get decent approximations for #sqrt(30)# by truncating just before one of the #10#'s as follows:

#sqrt(30) ~~ 5+1/2 = 11/2#

#sqrt(30) ~~ 5+1/(2+1/(10+1/2)) = 241/44#

#sqrt(30) ~~ 5+1/(2+1/(10+1/(2+1/(10+1/2)))) = 5291/966#

Let's stop there and use this to give us an approximation for #sqrt(270)#...

#sqrt(270) = 3sqrt(30) ~~ 3*5291/966 = 15873/966 ~~ 16.431677#