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

Nov 2, 2016

You can use the fact that if a positive number $a$ is that $a < x$, then ${a}^{2} < {x}^{2}$, and vice versa.
Then a number $a$ such that ${a}^{2} < 97$ means that $a < \sqrt{97}$. To solve the problem the number $a$ must also satisfy $97 < {\left(a + 1\right)}^{2}$.
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 $9 < \sqrt{97} < 10$