# Estimating Square roots

Question: Estimate the value to the nearest tenth

$$\sqrt{47}$$

But I don't know how I could estimate without using the calculator

Thank You and Help is appreciated

• From your name, GCSE, I suspect you just need to think that the answer is just below 7. Then guess and check by multiplying. – Karl May 27 at 16:55
• Specifically $x^2-47=0$ by trial and improvement. – Karl May 27 at 16:59
• $47=7^2-2$ leads to $\sqrt{47}=\sqrt{49-2}=7\sqrt{1-\frac{2}{49}}\approx 7\left(1-\frac{1}{49}\right)=7-\frac{1}{7}.$ – Jack D'Aurizio May 27 at 18:21

The $$\sqrt{47}$$ is between $$\sqrt{36}$$ and $$\sqrt{49}$$, so that means $$6<\sqrt{47}<7$$. Now, we can use casework. To find $$n^2=47$$, we first try a greater number, such as $$6.7$$. Using $$6.7$$, we find $$6.7^2=44.89$$, so $$6.7<\sqrt{47}$$. Next, we try $$6.9$$. $$6.9^2=47.61$$. We try $$6.8$$ for measure, and we find $$6.8^2=46.24$$. Therefore, $$\sqrt{47}$$ rounded to nearest tenth is $$6.9$$

Typically the binomial theorem is used in this type of situation as follows: $$\sqrt{47}=7\sqrt{\frac{47}{49}}=7\sqrt{1-\frac2{49}}$$ $$\therefore\sqrt{47}\approx7\left(1+\frac12\left(-\frac2{49}\right)\right)=\frac{48}7\approx6.9$$ With the final answer rounded to the nearest tenth. But all you actually need to do is find the values of some estimates squared and conclude the rounded value of the answer from these results. $$6.8^2=46.24$$ $$6.85^2=46.9225$$ $$6.9^2=47.61$$ So from this you can see that $$6.85^2\lt 47 \lt 6.9^2$$ So you can conclude that $$6.85\lt\sqrt{47}\lt6.9$$ So the value of $$\sqrt{47}$$ is $$6.9$$ to the nearest tenth.

• Note that the formula derived from the binomial formula here is the same as Heron's (a.k.a. Babylonian a.ka. Newton's) method for $x=7$, cf. my answer. If you had written $\sqrt{47} = 6\sqrt{1+\frac{11}{36}}$ instead, you would have gotten Heron's estimate for $x=6$. I personally think that Heron's method is a bit more basic than the binomial (a.k.a. Taylor) expansion of the square root. (Arguably, the student is supposed to do it with trial and error anyway.) – Torsten Schoeneberg May 27 at 17:27

When dealing with algebraic integers / numbers of degree two (radicals $$\sqrt N$$ basicly, $$N$$ an integer,) the (subjectively) best way to approximate by using rational numbers is to build the continued fraction of the number.

This answers goes in this direction. (Using the calculator, we get a priori some $$6.855654600\dots$$ value, and we have to consider at least the approximation $$6.855$$, then round to $$6.9$$.)

Let us compute the continued fraction of the given number $$a=\sqrt{47}$$:

\begin{aligned} a &= \boxed{6} + \color{blue}{(a-6)} \\ &= \boxed{6}+\frac{a+6}{a^2-6^2} = \boxed{6}+\frac1 {\displaystyle \frac{a+6}{11}} = \boxed{6}+\frac1 {\displaystyle \boxed{1}+\frac{a-5}{11}} \\ &= \boxed6+\frac1 {\displaystyle \boxed1 + \frac1 {\displaystyle \frac{11(a+5)}{a^2-5^2}}} = \boxed6+\frac1 {\displaystyle \boxed1 + \frac1 {\displaystyle \frac{a+5}{2}}} \\ &= \boxed6+\frac1 {\displaystyle \boxed1 + \frac1 {\displaystyle \boxed 5 + \frac{a-5}{2}}} \\ &= \boxed6+\frac1 {\displaystyle \boxed1 + \frac1 {\displaystyle \boxed 5 + \frac 1 {\displaystyle \boxed 1 +\frac {a-6}{11}}}} \\ &= \boxed6+\frac1 {\displaystyle \boxed1 + \frac1 {\displaystyle \boxed 5 + \frac 1 {\displaystyle \boxed 1 +\frac1 {\displaystyle \boxed{12} + \color{blue}{(a-6)}}}}} \\ &=\dots \end{aligned} Note that the $$\color{blue}{(a-6)}$$ was "already computed". So considering the calculus from the first blue position, and implementing it recursively on the last blue position we get better and better approximations. Continued fractions are collecting in this sense only the "boxed data", it is a good way to denote this number like: $$a =\sqrt{47} =[6;\ 1,5,1,12,\ 1,5,1,12\ ,\ \dots] =[6;\ \overline{1,5,1,12}]$$ Now the fractions obtained by truncation of this infinite representations are better and better approximations of the given number $$a$$. Let us see them: \begin{aligned} x_0 &= [6;] = 6\ ,\\ x_1 &= [6;1] = 6+1/1=7\ ,\\ x_2 &= [6;1,5] = 6+1/(1+1/5)=41/6=6.8(3)\dots\ ,\\ x_3 &= [6;1,5,1] = 6+1/(1+1/(5+1/1))=48/7=6.(857142)\ ,\\ x_3 &= [6;1,5,1,12] = 6+1/(1+1/(5+1/(1+1/12)))=617/90=6.8(5)\ ,\\ \end{aligned} and so on. Observe that the "convergents" above are respectively less, more, less, more, less, ... than $$a=\sqrt{47}$$. At this point we can enclose $$a$$ between the numbers - 6 . 857142857142... and - 6 . 855555555555... so the best tenth approximating is obtained by rounding up to $$6.9$$.

In fact, some computer algebra software like sage computes easily something like:

sage: K.<a> = QuadraticField(47)
sage: c = a.continued_fraction()
sage: c
[6; (1, 5, 1, 12)*]

sage: c.convergents()
lazy list [6, 7, 41/6, ...]

sage: for k in [0..10]:
....:     ck = c.convergent(k)
....:     print "%s. convergent = %s = %s" % (k, ck, ck.n())
....:
0. convergent = 6 = 6.00000000000000
1. convergent = 7 = 7.00000000000000
2. convergent = 41/6 = 6.83333333333333
3. convergent = 48/7 = 6.85714285714286
4. convergent = 617/90 = 6.85555555555556
5. convergent = 665/97 = 6.85567010309278
6. convergent = 3942/575 = 6.85565217391304
7. convergent = 4607/672 = 6.85565476190476
8. convergent = 59226/8639 = 6.85565458965158
9. convergent = 63833/9311 = 6.85565460208356
10. convergent = 378391/55194 = 6.85565460013770


(The above $$10$$.th convergent is for instance the best fraction approximation of $$a$$ with denominators $$\le 55194$$. The bigger coefficients appear in the continued fraction, the better approximation, and the less needed $$k$$ to get a given bound. In our case, each time we pass through the $$12$$ in the continued fraction, we get a "jumped better approximation", well, just my feeling...)

• I enjoyed your answer +1 – Karl May 27 at 17:59

Let $$f(x) = y = \sqrt x$$

At $$x=49$$ ,$$y=\sqrt{49} = 7$$

Let $$dx\approx\Delta x = -2$$ so that $$x+\Delta x = 49-2=47$$

Now, $$dy = \frac{1}{2\sqrt x}dx = \frac{1}{2\sqrt x}dx =-2\frac{1}{2\sqrt{49}}\approx-0.1428\cdots$$

$$\Delta y \approx dy = - 0.1428\cdots$$

So, $$f(x+\Delta x) = f(47) = y +\Delta y \approx 7-0.14 \approx 6.86$$

$$\sqrt47 \approx 6.86 \ or \ \sqrt{47} \approx 6.9$$

• (+1) nice approach but I assume that the OP wouldn't understand what a derivative is at GCSE level. – Peter Foreman May 27 at 17:11
• Thank you @PeterForeman, but here in India they teach that in the age group of 16 - 17, and +1 for your answer too 😄 – Ak19 May 27 at 17:16
• Yes but GCSE students are 15-16 when they take the exam and younger when learning. – Peter Foreman May 27 at 17:19
• Sorry, I don't know about that – Ak19 May 27 at 17:20

This is how to do it without a calculator. We will show in detail how to estimate the square root of $$857.142$$.

1. "Going out" left and right from the decimal point, break the number up into pairs: $$857.142 \to 8 \ 57. \ 14 \ 20$$.
2. Write this out as you would a division problem. (We will only sort of treat it like one.) $$\begin{array}{r} & &-- &-- &. &-- &-- \\ &) & 8 &57 &. &14 &20 \end{array}$$
3. Find the largest integer,$$n$$, such that $$n^2 \le 8$$. That would be $$n=2$$ Place $$n=2$$ and subtract $$n^2=4$$ from $$8$$ as is shown below. Also bring down the next two digits as is shown below. $$\begin{array}{r} & & 2 \\ & &-- &-- &. &-- &-- \\ &) & 8 &57 &. &14 &20 \\ 2 & & 4 \\ & &-- \\ & & 4 &57\\ \end{array}$$
4. This is easy to do, but tricky to explain. So far, your answer is $$2$$. Double that number, $$2\times 2 = 4$$, and $$\color{red}{\text{imagine}}$$ a digit $$\color{red}{\text{n}}$$ in the places shown. $$\begin{array}{r} & & 2 &\color{red}{\text{n}}\\ & &-- &-- &. &-- &-- \\ &) & 8 &57 &. &14 &20 \\ 2 & & 4 \\ & &-- \\ & & 4 &57\\ 4\color{red}{\text{n}} \\ & &-- &-- \\ \end{array}$$
5. Your task is to find the largest digit, $$\color{red}{\text{n}}$$, such that $$\color{red}{\text{n}} \times 4\color{red}{\text{n}} \le 457$$. In this case, $$4\color{red}{\text{n}} = 9$$. Replace the $$4\color{red}{\text{n}}s$$ with $$9s$$ and subtrace out $$9 \times 49 = 441$$ as shown. Also bring down the next two digits. $$\begin{array}{r} & & 2 & 9\\ & &-- &-- &. &-- &-- \\ &) & 8 &57 &. &14 &20 \\ 2 & & 4 \\ & &-- \\ & & 4 &57\\ 49 & & 4 &41\\ & &-- &-- \\ & & &16 &. &14 \end{array}$$
6. Doing just as before, the next step looks like this. $$\begin{array}{r} & & 2 & 9 &. &\color{red}{\text{n}}\\ & &-- &-- &. &-- &-- \\ &) & 8 &57 &. &14 &20 \\ 2 & & 4 \\ & &-- \\ & & 4 &57\\ 49 & & 4 &41\\ & &-- &-- \\ & & &16 &. &14 \\ 58\color{red}{\text{n}} \\ & & &-- &-- \\ \end{array}$$
7. Since $$582\times2 = 1164$$ and $$583\times3 = 1749$$, then $$\color{red}{\text{n}}=2$$ $$\begin{array}{r} & & 2 & 9 &. & 2\\ & &-- &-- &. &-- &-- \\ &) & 8 &57 &. &14 &20 \\ 2 & & 4 \\ & &-- \\ & & 4 &57\\ 49 & & 4 &41\\ & &-- &-- \\ & & &16 &. &14 \\ 582 & & &11 &. &64 \\ & & &-- &. &-- \\ & & & 4 &. &50 \\ \end{array}$$
8. This is the finished problem. Note that by adding $$00$$'s after the $$20$$, a more accurate answer can be obtained. $$\begin{array}{r} & & 2 & 9 &. & 2 & 7 \\ & &-- &-- &. &-- &-- \\ &) & 8 &57 &. &14 &20 \\ 2 & & 4 \\ & &-- \\ & & 4 &57\\ 49 & & 4 &41\\ & &-- &-- \\ & & &16 &. &14 \\ 582 & & &11 &. &64 \\ & & &-- &. &-- \\ & & & 4 &. &50 &20\\ 5847 & & & 4 &. &09 &29 \\ & & &-- &. &-- &--\\ & & & &. &40 &91 \\ \end{array}$$

## Applying the same process to $$47$$, we get

$$\begin{array}{r} & & 6 &. & 8 & 5 & 5 \\ & &-- &. &-- &-- &--\\ &) &47 &. &00 &00 &00\\ 6 & &36 &.\\ & &-- \\ & &11 &. &00\\ 128 & &10 &. &24\\ & &-- &. &-- \\ & & &. &76 &00 \\ 1365 & & &. &68 &25 \\ & & &. &-- &-- \\ & & &. & 7 &75 &00\\ 13705 & & &. & 6 &85 &25 \\ & & &. &-- &-- &-- \\ & & &. & &89 &75\\ \end{array}$$

To see whether or not to round off the last digit, try $$5$$ for the next digit. We would get $$137105 \times 5 = 685525 < 897500$$. So $$6.856$$ is closer to the true answer.

• (+1) Even though this answer was posted $7$ minutes after mine, I think it's obvious that you were working on it before I posted. – robjohn May 28 at 12:52
• @robjohn Thanks. They used to teach this algorithm in high school. It seems that many arithmetic skills have gone the way of cursive writing. – steven gregory May 28 at 14:15

Macluarin series of $$\sqrt{49-x}$$ is $$\sqrt{49-x}=7-\frac{x}{14}-\frac{x^2}{2744}-...$$ at $$x=2$$ $$\sqrt{47}\approx 7-\frac{1}{7}$$

There is also the Scaffold Square Root Algorithm (Digit by Digit with Examples): $$\require{enclose} \begin{array}{rl} \color{#090}{6}.\phantom{0}\color{#090}{8}\,\phantom{0}\color{#090}{5}\,\phantom{0}\color{#090}{5}\\[-4pt] \enclose{radical}{47.00\,00\,00}\\[-4pt] \underline{36}\phantom{.00\,00\,00}&\quad\leftarrow\color{#090}{6}\cdot\color{#090}{6}\\[-4pt] 11\,00\phantom{\,00\,00}\\[-4pt] \underline{10\,24}\phantom{\,00\,00}&\quad\leftarrow\color{#C00}{12}\color{#090}{8}\cdot\color{#090}{8}\\[-4pt] 76\,00\phantom{\,00}\\[-4pt] \underline{68\,25}\phantom{\,00}&\quad\leftarrow\color{#C00}{136}\color{#090}{5}\cdot\color{#090}{5}\\[-4pt] 7\,75\,00\\[-4pt] \underline{6\,85\,25}&\quad\leftarrow\color{#C00}{1370}\color{#090}{5}\cdot\color{#090}{5}\\[-4pt] \phantom{0.00}\,89\,75 \end{array}$$ The digits in red are twice the digits collected so far on top of the vinculum.

• I first learned this algorithm from a book about abacus calculations, so I believe it is probably pretty old. – robjohn May 27 at 19:51

There are many methods to compute square roots of a number $$s$$. One that predates calculators by several millenia is Heron's method a.k.a. the Babylonian method (which also is an easy case of Newton's method), and it roughly says that if you have first not-too-bad estimate $$x$$, then

$$\dfrac12 (x + \dfrac{s}{x})$$

-- the average of $$x$$ and $$s/x$$ -- will be an even better estimate for $$\sqrt s$$.

For $$x=6$$, the calculation is easily done without a calculator and gives $$\frac{83}{12} = 6 \frac{11}{12}$$.

For $$x=7$$, the calculation is easily done without a calculator and gives $$\frac{48}{7} = 6 \frac{6}{7}$$.

Note that both are $$\approx 6.9$$. You can of course take either value and put in the formula again to get better estimates.