How do you find two consecutive whole numbers that #sqrt97# lies between?

1 Answer
Nov 2, 2016

Answer:

You can use the fact that if a positive number #a# is that #a < x#, then #a^2 < x^2#, and vice versa.

Explanation:

Then a number #a# such that #a^2<97# means that #a < sqrt97#. To solve the problem the number #a# must also satisfy #97 < (a+1)^2#.

But the number #a# must be #a < 10#, since #10^2 =100 > 97#. If we then try 9, we find out that #9^2 = 81 < 97#.

The answer is then #9# and #10#, because #9^2 < 97 < 10^2#, and this means #9 < sqrt97 < 10#