In which of the intervals is $\sqrt{12}$

Question

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!

Answer 1

$$\left(\frac{7}2 \right)^2 = \frac{49}{4}>12>3^2$$

Hence the solution is $B$.

Answer 2

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 $$3 < \sqrt{12} < 3.5.$$

Answer 3

$$\sqrt{12} = 3 + (\sqrt{12} - 3) = 3 + \frac{(\sqrt{12})^2 - 3^2}{\sqrt{12} + 3} = 3+\frac{3}{\sqrt{12}+3}$$

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.

Answer 4

Building upon what Siong already said. For numbers greater than one, the square function is an increasing function. This means that if

$$a>b$$

then

$$a^2>b^2$$

eg

$$5<6$$

$$25<36$$

(this doesn't apply for numbers less than 1, you can try yourself)

Applying the same logic but in reverse, find squares of all options and figure out in which interval you will find 12. Taking square root will give you the original options and $sqrt12$ in the interval

Answer 5

$3^2=9.$

$3.5^2 = \frac{35}{10} \times \frac{35}{10} = \frac{35 \times 35}{100} = \frac{1225}{100} = 12.25.$

$9 < 12 < 12.25.$

$\sqrt{9} < \sqrt{12} < \sqrt{12.25}.$

$3< \sqrt{12} < 3.5.$

Answer 6

There is still a faster way :)

Tool: Newton's well-known formula

If $\sqrt a=n, ~n \in\mathbb R^{+}\setminus \mathbb Q^{+}$ and $\lfloor \sqrt a \rfloor=m$, then

$$\sqrt a <\frac{ \dfrac am +m}{2}≈\sqrt a$$

So, you can get the answer right away:

$$\sqrt {12}<\frac{ \dfrac {12}{3} +3}{2}=3.5$$