In which of the intervals is $\sqrt{12}$ In which of the intervals is $\sqrt{12}:$
a) $(2.5;3);$
b) $(3;3.5);$
c) $(3.5;4);$
d)
$(4;4.5)$?
We can use a calculator and find that $\sqrt{12}\approx3.46$ so the correct answer is actually b. How can we think about the problem without a calculator (if on exam for example)? I was able to conclude that $$\sqrt{9}=3<\sqrt{12}<\sqrt{16}=4,\\3<\sqrt{12}<4,\\\sqrt{12}\in\left(3;4\right).$$ How can I further constrict the interval? Thank you in advance!

The smallest integer square greater than $12$ is $4^2 = 16$, and the largest integer square less than $12$ is $3^2 = 9$.  Therefore, $3 < \sqrt{12} < 4$.  This eliminates (a) and (d) as choices.  The next thing to do is consider the value of $(3.5)^2$, which is $$(3 + 1/2)^2 = 9 + 2(3)(1/2) + (1/2)^2 = 12 + 1/4 > 12.$$  Hence $$3 < \sqrt{12} < 3.5.$$
