Rough root

Answer: 2.2

Squaring the options

$\dfrac{10 000}{2012}\approx\dfrac{10000}{2000}=5$, so $\sqrt{\dfrac{10000}{2012}}\approx\sqrt5$

(We should really check how good this approximation is. See below (or click here) for comparison).

So one of the options should square to a number close to $5$. Squaring them gives:

$1.9^2=3.61$ which is too small

$2.2^2=4.84$ which is too small

$2.5^2=6.25$ which is too big

so $2.7^2$ will definitely be too big.

$2.2$ is too small and $2.5$ is too big. To find which is closer to $\sqrt5$, test a number in between.

If $2.3$ is too big, then we will know that $\sqrt5$ is between $2.2$ and $2.3$, so it is definitely closer to $2.2$ than to $2.5$.

$2.3^2=2.3\times2+2.3\times0.3=4.6+0.69=5.29$, which is too big.

So $2.2$ is the closest.

Estimating the square root

$10 000=100^2$

$\dfrac{100}?\times\dfrac{100}?=\dfrac{10000}{?\times?}$ where $?\times?$ is close to $2012$

$40^2=1600$ and $50^2=2500$. Try $45^2=2025$

$\therefore\dfrac{100}{45}\times\dfrac{100}{45}=\dfrac{10000}{2025}\approx\dfrac{10000}{2012}$

(We should really check how close this approximation is. See below (or click here) for comparison)

So $\sqrt{\dfrac{10000}{2012}}\approx\dfrac{100}{45}=\dfrac{20}{9}=2.\dot2$

So $2.2$ is the closest.

How close are our approximations?

The difference between $\dfrac{10 000}{2012}$ and $5$ is: $$\begin{array}{rl}\frac{10000}{2000}-\frac{10000}{2012}& =1000(\frac{1}{2000}-\frac{1}{2012})\\ & =10000\left(\frac{2012-2000}{2000\times 2012}\right)\\ & =\frac{10000\times 12}{2000\times 2012}\\ & =\frac{10\times 6}{2012}=\frac{30}{1006}<\frac{30}{1000}=0.03\end{array}$$ So really we were looking for a number whose square was somewhere between $5$ and $5.03$. Squaring the options gave results far less precise than this, so this approximation was good enough for this situation.

The difference between $\dfrac{10 000}{2012}$ and $\dfrac{10000}{2025}$ is: $$\begin{array}{rl}{\displaystyle \frac{10000}{2012}}-{\displaystyle \frac{10000}{2025}}& =10000({\displaystyle \frac{1}{2012}}-{\displaystyle \frac{1}{2025}})\\ & =10000\left(\frac{2025-2012}{2012\times 2025}\right)\\ & =\frac{10000\times 13}{2025\times 2012}<\frac{10000\times 13}{2000\times 2000}=\frac{13}{2\times 200}<\frac{16}{400}=\frac{4}{100}=0.04\end{array}$$

So the square of $2.\dot2$ is too large, by up to $0.04$.

We can write this as $2.\dot2^2-c^2<0.04$, where $c$ is the exact value of $\sqrt{\dfrac{10000}{2012}}$

We can factorise $2.\dot2^2-c^2$ as the difference of two squares:

$2.\dot2^2-c^2=(2.\dot2+c)(2.\dot2-c)$, so $(2.\dot2+c)(2.\dot2-c)<0.04$

$2.\dot2-c$ must be very close to $0$, because the product is close to $0$, and $(2.\dot2+c)$ is not particularly close to $0$. This is good, because if $2.\dot2$ is a good approximation, then $2.\dot2-c$ is close to $0$.

In fact, $(2.\dot2+c)$ must be more than $2$, since $c$ is positive. So $2.\dot2-c$ must less than $0.02$, to give a product of less than $0.04$ (since $2\times0.02=0.04$).

So $2.\dot2$ is an over-estimation by less than $0.02$, which means that the true value of $c$ is somewhere between $2.\dot2-0.02=2.20\dot2$ and $2.\dot2$. So $2.2$ is defiitely the closest of the options. Our approximation was actually far better than the question required.