# Math Help - Calculate the square root of 65 by hand

1. ## Calculate the square root of 65 by hand

Can someone please show me how to calculate the square root of 65 by hand, thanks.

2. $\mbox{Let} \ x_0=\sqrt{65} \ \mbox{then} \ a_0=\left\lfloor\sqrt{65}\right\rfloor}=8$

$\displaystyle a_k=\left\lfloor x_k\right\rfloor, \ \ x_{k+1}=\frac{1}{x_k-a_k}$

$\displaystyle x_1=\frac{1}{x_0-a_0}=\frac{1}{\sqrt{65}-8}\frac{\sqrt{65}+8}{\sqrt{65}+8}=\sqrt{65}+8 \Rightarrow\ a_1=\left\lfloor x_1\right\rfloor=16$

$\displaystyle x_2=\frac{1}{\sqrt{65}+8-16}=\frac{1}{\sqrt{65}-8}\frac{\sqrt{65}+8}{\sqrt{65}+8}=\sqrt{65}+8} \Rightarrow \ a_2=\left\lfloor x_1\right\rfloor=16$

$\sqrt{65}=[8;16,16,\cdots]=8+\frac{1}{16+\frac{1}{16+\frac{1}{\cdots}}}$

3. Hi would you also be able to show the long division method? Thanks.

4. I am not familiar with that method.

Can you provide an example?

5. By the way, I noticed there was code in this file, what language or program is that? Or is part of the Web site?

6. Here is a link that shows that method but since for 8 is the first number, I am not sure how to place the double of that number.

By the way, I noticed there was code in this file, what language or program is that? Or is part of the Web site?
Part of the website.

you type math in [] and \math in [].

Then in between [] here [\] you type in latex.

Helpisplaying a formula - Wikipedia, the free encyclopedia

LaTeX Online Equation Editor

8. Just do division as it is shown.

It is sort of like the recursion formula.

Can someone please show me how to calculate the square root of 65 by hand, thanks.
Since 65 is so close to 64 (a perfect square), I think that the intended method for doing this question is to use the binomial expansion $(1+x)^{1/2} = 1+\frac12x-\frac18x^2+\ldots$.

You can write $\sqrt{65}$ as $\sqrt{64+1} = 8\sqrt{1+\frac1{64}$. Then use the above binomial series with $x = 1/64$ to estimate the square root. Because 1/64 is such a small number, its powers decrease very rapidly, and you only need to take a few terms of the series to get an accurate approximation to the square root.

10. First, you need to know an important property of square roots...

$\displaystyle (\sqrt{a})^2 = a$

$\displaystyle \sqrt{a}\sqrt{a} = a$

$\displaystyle \sqrt{a} = \frac{a}{\sqrt{a}}$.

In other words, if you divide a number by its square root, you get the square root.

The standard method is to use the Babylonian Method of "Guess, Divide, Check, Average".

First, note that $\displaystyle 64 < 65 < 81$

$\displaystyle \sqrt{64}<\sqrt{65}<\sqrt{81}$

$\displaystyle 8<\sqrt{65}<9$.

A good initial guess is the average of the endpoints, so $\displaystyle \frac{8+9}{2} = 8.5$.

Now we divide $\displaystyle \frac{65}{8.5} = 7.6470588235\dots$.

Clearly, our square root should be less than $\displaystyle 8.5$. By averaging our guess and our quotient, we get a new guess.

$\displaystyle \frac{8.5 + 7.6470588235\dots}{2} = 8.07352941176\dots$.

Now start again with $\displaystyle 8.07352941176$ as your guess. Continue until you have the desired level of accuracy.

11. Newton's method for numerically solving equations, applied to the square root function, gives his:

It is clear that 8 is close to the square root of 65 since $8^2= 64$

65/8= 8.125. Notice that since 8 is a little smaller than the square root of 65, 65/8 is a little larger. The correct square root is somewhere between them. Try half way: (8+ 8.125)/2= 16.125/2= 8.0625.

Now repeat: 65/8.0625= 8.06201. Now, since 8.0625 was just a little larger than the square root of 8, 65 divided by 8.0625 is just a little smaller. The correct square root is somewhere between 8.0625 and 8.06201. Notice that we already know that, to three decimal places, the square root of 65 is 8.062 but we can continue this, getting a more and more accurate answer, as long as we can do that division by hand.

12. Using calculus

$\displaystyle f(x+h)\approx f(x)+h\times f'(x)+\dots$

$\displaystyle (x+h)^{\frac{1}{2}}\approx x^{\frac{1}{2}}+h\times {\frac{1}{2}}x^{\frac{-1}{2}}+\dots$

$\displaystyle (64+1)^{\frac{1}{2}}\approx 64^{\frac{1}{2}}+1\times {\frac{1}{2}}64^{\frac{-1}{2}}+\dots$

13. Using a ruler...

Draw a right-angled triangle,
perpendicular sides 8 units and 1 unit.
Measure the hypotenuse.

That's the closest I can get to "by hand".

14. Originally Posted by Archie Meade
Using a ruler...

Draw a right-angled triangle,
perpendicular sides 8 units and 1 unit.
Measure the hypotenuse.

That's the closest I can get to "by hand".
I'm sure when the OP says "by hand" he/she means "without using a square root function on a calculator".

But I do like your idea

15. ## Re: Calculate the square root of 65 by hand

(I'm not very good at the latex stuff, but there is an algorithm that you can use to provide an answer to any number of digits of accuracy.)

The number must be grouped first into two digit segments from the decimal point, so for 12345.6789 we have 1 23 45. 67 89. The left most digit pair in this case 1 (which is in effect 01, since any number of leading zeros won't affect the number value) . Thus we first find the root closest that is less than or equal to that pair of digits, so it is 1.

For our number 65, the closest square which is less than or equal to 65 is 64, or eight squared.

We deduct this from 65, which leaves 1.

Since the square root of 65 is greater than 8, we must make this integer number a decimal number 65.00 00 00 00

==8. 0 6 2 2
----------------
+65.00 00 00 00
-64 This is 8 squared, subtract this from our number.
----
=01 00 Bring down first decimal pair of zero digits 00
=01 00 00 The 8 is doubled to give 16, and we must find a value 16_x_ <= 100, as we can't, we write down 0 above and bring down next 00 digits
- 96 36 The zero was doubled but gives zero anyway, so the inequality becomes 160_x_ <= 10000, we find 6 satisfies such that 1606 x 6 = 9636
--------- Subtract
= 3 64 00 and bring down next 00 pair. We double the 6 and add it to our inequality (carry the one) such that 16[0+1]2_x_<=36400, or 1612_x_<=36400
- 3 22 44 We find 2 satisfies this inequality, such that 16122x2 = 32244 <= 36400
------------ Subtract and bring down next 00 pair
= 41 56 00 Now the 2 is doubled and added to the inequality 16124_x_ <= 415600,
- 32 24 84 161242x2 = 322484 <= 415600,
---------------
= 9 31 16

You could go on indefinitely depending on the level of accuracy you want. Remember, every new digit of the root you discover, you must change the inequality to append it on the right of that number with a new mystery number "_" to be found. When the double it and it is 10 or over, you must carry the one to the next column.

My photo isn't very clear due to the fuzzy webcam.

Hope this is what you're looking for.

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