Question

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

Answer 1

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`