Find two consecutive whole numbers between which #sqrt71# lies.?

2 Answers
Feb 21, 2018

The two numbers are #8# and #9#.

Explanation:

To figure this out, first, I would list a few numbers and their radical equivalents. I'll start from #6#:

#color(white){color(black)( (6, qquadsqrt36), (7, qquadsqrt49), (8, qquadsqrt64), (9, qquadsqrt81), (10, qquadsqrt100)):}#

Now, we have to find the two radicals that are consecutive where one of them is bigger than #sqrt71# and the other is smaller.

These two numbers are #sqrt64# and #sqrt81# (because #64<=71<=81#). That means that the two whole numbers are #8# and #9#.

You can verify this by using a calculator to compute #sqrt71#:

#sqrt71~~8.42614...#

This number is between #8# and #9#, so our answer is correct.

Feb 21, 2018

#8# and #9#

Explanation:

To find two consecutive whole numbers that #sqrt71# lies between, we must first find two square roots that a) are consecutive perfect squares (are the squares of whole numbers) and b) #sqrt(71)# lies between.

Let's write out some perfect squares:

#1^2=1#
#2^2=4#
#3^2=9#
#4^2=16#
#5^2=25#
#6^2=36#
#7^2=49#
#8^2=64#
#9^2=81#

From the above, we see that #sqrt(64)# and #sqrt(81)# meet our above conditions: they are consecutive perfect squares and #sqrt(71)# lies between them.

So, #sqrt(71)# lies between #sqrt(64)=8# and #sqrt(81)=9#