Find the square root of $404.11$.

Question

Find the square root of $404.11$ without using calculator accurate upto $2$ decimal places .

It is clear that

$20<\sqrt{404.11}<21$ so it will be $20.ab$

without trial and error what could be the fast way to compute it .

Answer 1

This is the traditional way finding square roots was taught a long time ago.

With integer $0 \leq a \leq 9,$ $$ \left(20 + \frac{a}{10}\right)^2 = 400 + 4 a + \frac{a^2}{100} $$ We see that $a=1 $ is the largest possible choice, so we have reached $20.1$

now $20.1^2 = 404.01$

Next, With integer $0 \leq b \leq 9,$ $$ \left(20.1 + \frac{b}{100}\right)^2 = 404.01 + \frac{40.2 b}{100} + \frac{b^2}{10000} $$

We are trying to reach 404.11. We have only $1/10$ to go. It follows that $40.2 b < 10$ and $b = 0.$

STILL TYPING!!!!!!!! i PUT IN A BIT MORE AND SAVE!!!!!!!!!!!

Next, With integer $0 \leq c \leq 9,$ $$ \left(20.10 + \frac{c}{1000}\right)^2 = 404.01 + \frac{40.2 c}{1000} + \frac{c^2}{1000000} $$

We are trying to reach 404.11. We have $1/10$ to go. It follows that $40.2 c < 100$ and we may take $c = 2.$

Very good approximation, and not too lartge, with $20.102$

Indeed, with $c=2,$ we check $$ \color{blue}{ \left(20.10 + \frac{2}{1000}\right)^2 = 404.01 + \frac{80.4 }{1000} + \frac{4}{1000000} = 404.01 + 0.0804 + 0.000004 = 404.090404 } $$

Furthermore, if we had taken $c$ one larger we would exceed the target, as $$ \left(20.10 + \frac{3}{1000}\right)^2 = 404.01 + \frac{120.6 }{1000} + \frac{9}{1000000} = 404.01 + 0.1206 + 0.000009 = 404.130609 $$ which is too large.

$$\begin{array}{l}\text{Little Jack Horner}\cr \text{Sits in a corner}\cr \text{Extracting cube roots to infinity,}\cr \text{An assignment for boys}\cr \text{That will minimize noise}\cr \text{And produce a more peaceful vicinity.}\end{array}$$
This is verse number 17 in The Space Child's Mother Goose published in 1958, verses by Frederick Winsor, illustrations by Marian Parry. The method shown above, modified only slightly, works just as well for cube roots.

Answer 2

You can also use the tangent line approximation , that holds for h " small-enough"

$$f(x+h) \approx f(x)+hf'(x)$$

With $x=400, h=4.11 f(x)= \sqrt x$. Then $f'(x)=\frac {1}{2 \sqrt x}$

Then you get : $$ \sqrt {404.11} \approx 20+ (4.11) \frac {1}{40}=20.10275 $$

Which is not a bad approximation.

And you cannot 'compute it', since the square root is irrational.

Answer 3

Mental arithmetic approximation:

Clearly $200^2=40000$ and $201^2=40401$ and $202^2=40804$.

So interpolating $\frac{40411-40401}{40804-40401} = \frac{10}{403} \approx \frac1{40}$, I would estimate that $201.025^2$ is about $40411$ and so $$\sqrt{404.11} \approx 20.1025.$$

Answer 4

There's a long-division-looking way to calculate square roots that I learned from a book when I was a kid. It's not used very often, as far as I can tell, although I used it to calculate the square root of $2$ to unreasonable precision (for hand calculation, anyway).

I'm not sure how easy it will be to do this in LaTeX, but here goes.

You first write the square root with the digits grouped in pairs, going outward from the decimal point:

$$ \begin{align} & \sqrt{4 \; 04. \; 11 \; 00 \; 00} \end{align} $$

Give the floor of the square root of the first grouping; in this case, $\sqrt{4} = 2$ exactly, so write $2$:

$$ \begin{align} & \;\,\, 2 \\ & \sqrt{4 \; 04. \; 11 \; 00 \; 00} \end{align} $$

The square of $2$ is $4$, so write $4$ underneath, and then subtract:

$$ \begin{align} & \;\,\, 2 \\ & \sqrt{4 \; 04. \; 11 \; 00 \; 00} \\ & \;\,\, \underline{4 \; \phantom{00}} \\ & \;\,\, \phantom{4 \; 0}4 \end{align} $$

Now, at left, you double your current root estimate to get $4$, and ask what the largest digit x is for which $4$x (a two-digit number, not $4$ times x) times x is less than $4$:

$$ \begin{align} & \;\,\, 2 \; \phantom{0}\mbox{x}. \\ & \sqrt{4 \; 04. \; 11 \; 00 \; 00} \\ & \;\,\, \underline{4 \; \phantom{00}} \\ & \;\,\, \phantom{4 \; 0}4 \\ 4\mbox{x} & \end{align} $$

Obviously, the answer is $0$, so we write $0$ on top, subtract, and drop down another pair of digits:

$$ \begin{align} & \;\,\, 2 \; \phantom{0}0. \\ & \sqrt{4 \; 04. \; 11 \; 00 \; 00} \\ & \;\,\, \underline{4 \; \phantom{00}} \\ & \;\,\, \phantom{4 \; 0}4 \\ 40 & \;\,\, \phantom{4 \; 0}\underline{0 \; \phantom{11}} \\ & \;\,\, \phantom{4 \; 0}4. \; 11 \end{align} $$

Again we double our current estimate to get $40$, and ask what the largest digit x is for which the three-digit number $40$x times x is less than $411$:

$$ \begin{align} & \;\,\, 2 \; \phantom{0}0. \; \phantom{1}\mbox{x} \\ & \sqrt{4 \; 04. \; 11 \; 00 \; 00} \\ & \;\,\, \underline{4 \; \phantom{00}} \\ & \;\,\, \phantom{4 \; 0}4 \\ 40 & \;\,\, \phantom{4 \; 0}\underline{0 \; \phantom{11}} \\ & \;\,\, \phantom{4 \; 0}4. \; 11 \\ 40\mbox{x} \end{align} $$

This time, $401$ times $1$ equals $401$, which is less than $411$, so we write:

$$ \begin{align} & \;\,\, 2 \; \phantom{0}0. \; \phantom{1}1 \\ & \sqrt{4 \; 04. \; 11 \; 00 \; 00} \\ & \;\,\, \underline{4 \; \phantom{00}} \\ & \;\,\, \phantom{4 \; 0}4 \\ 40 & \;\,\, \phantom{4 \; 0}\underline{0 \; \phantom{11}} \\ & \;\,\, \phantom{4 \; 0}4. \; 11 \\ 401 & \;\,\, \phantom{4 \; 0}\underline{4. \; 01} \\ & \;\,\, \phantom{4 \; 04.} \; 10 \; 00 \end{align} $$

We continue on, always doubling the current estimate and writing it on the left, and then affixing an x at the end, and also at the end of the current estimate. We then ask, at each iteration, what the largest $x$ for which the left-hand number times the current estimate (including the x in both cases) is less than the "remainder"). The final tally:

$$ \begin{align} & \;\,\, 2 \; \phantom{0}0. \; \phantom{1}1 \; \phantom{0}0 \; \phantom{0}2 \\ & \sqrt{4 \; 04. \; 11 \; 00 \; 00} \\ & \;\,\, \underline{4 \; \phantom{00}} \\ & \;\,\, \phantom{4 \; 0}4 \\ 40 & \;\,\, \phantom{4 \; 0}\underline{0 \; \phantom{11}} \\ & \;\,\, \phantom{4 \; 0}4. \; 11 \\ 401 & \;\,\, \phantom{4 \; 0}\underline{4. \; 01} \\ & \;\,\, \phantom{4 \; 04.} \; 10 \; 00 \\ 4020 & \;\,\, \phantom{4 \; 04. \;} \underline{\phantom{10 \; 0}0 \; \phantom{00}} \\ & \;\,\, \phantom{4 \; 04. \;} 10 \; 00 \; 00 \\ 40202 & \;\,\, \phantom{4 \; 04. \; 1}\underline{8 \; 04 \; 04} \\ & \;\,\, \phantom{4 \; 04. \; 1}1 \; 95 \; 96 \end{align} $$

The "remainder" is a little less than half the left-hand number, so we can estimate further that the square root is a little less than $20.1025$. This process can be carried out ad nauseam.

It's a little involved, so if you have any questions, feel free to ask.

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ETA (2024-03-05): Incidentally, the original question can be answered by inspection if you remember that $21^2 = 441$, so similarly $20.1^2 = 404.01$, which leaves a residue of $0.1$. Since the derivative of $x^2$ near $x = 20$ is about $40$, that means we need to tack on an extra $0.1/40 = 0.0025$ approximately, giving $20.1025$. That is within $0.000013$ of the correct value.