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

Question

9343 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-02T21:55:48

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 $a < sqrt97$ . 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 $9 < sqrt97 < 10$

How do you find two consecutive whole numbers that sqrt97sqrt97 lies between?