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
 A: 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.
A: 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 $6.9$
A: 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, $f(x+\Delta x) = f(47) = y +\Delta y \approx 7-0.14 \approx 6.86$

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

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