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!

$$\sqrt{12} = 3 + (\sqrt{12} - 3) = 3 + \frac{(\sqrt{12})^2 - 3^2}{\sqrt{12} + 3} = 3+\frac{3}{\sqrt{12}+3}$$
and since $3 < \sqrt{12} < 4$, $\frac{3}{3 + 4} < \frac{3}{\sqrt{12} + 3} < \frac{3}{3 + 3}$. Thus $\sqrt{12} < 3 + \frac{1}{2}$ so $b$ is correct.
