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!

$3^2=9.$
$3.5^2 = \frac{35}{10} \times \frac{35}{10} = \frac{35 \times 35}{100} = \frac{1225}{100} = 12.25.$
$9 < 12 < 12.25.$
$\sqrt{9} < \sqrt{12} < \sqrt{12.25}.$
$3< \sqrt{12} < 3.5.$
