Arithmetic, Part 1 · Section 38
Extraction of Roots.
ANd firſt for the ſquare Root, that is, a ſquare Number being given, to find the root or ſide of it in a Number: which root or ſide, being multiplyed into it ſelf, muſt therefore produce that ſquare Number.
The doing of this is ſhewed in (almoſt) all books of Arithmetick; and the reaſon of it (in ſome) which is taken from the fourth propoſition of the ſecond book of Euclide; which ſaith,
If a right line be divided by chance; the ſquares made of the parts, together with the Rectangle made of the parts twice, is equal to the ſquare of the whole.
Example.
Now let AB, be ſuppoſed 20, AE 15, and BE 5.
Then the ſquare of AB, (which is made by multiplying the root 20 into it ſelfe) is equal to 400.
| And the ſquare AE, that is, 15 times 15 is equal to | 225 |
| And the ſquare of BE is 5 times 5 | 25 |
| And the rectangle AO, 5 times 15 | 75 |
| And the rectangle DO, 5 times 15 | 75 |
| In all | 400 |
which is equal to the ſquare of AB, as before.
If therefore a ſquare Number, as 334084 be given to have the root extracted: firſt, make a point over 4, (the place of Unity) and another over o, the ſecond figure from it; and ſo another over o, the ſecond figure from o, towards the left hand, obſerving the like order ſtill, if there were more ſecondary figures, point them all as in the Margine;
Then look at the figures under the firſt point towards the left hand which are 33, … and take the greateſt ſquare root of them, which is 5, (for 6 times 6 is 36) and ſay 5 times 5 is 25, which take out of of 33, there remaines …
Then multiply the root 5 by 20, it gives 100 for a Diviſor; and the Dividend is 840, namely, the figures which reach to the next point, and ſo the quotient might be 8, but muſt be but 7, becauſe the ſquare of the quotient, being now 49, muſt (together with 7 times the Diviſor) be taken out of 840.
And the remainer will be …
Adde this laſt quotient 7, being put to the former quotient 5, after the manner uſed in plain diviſion, it will be 57, which multiply ſtill by 20, the product is 1140 for a new Diviſor, and the Dividend (becauſe there remaines but one point) is all the figures 9184.
| And the quotient can be but 8, and muſt be ſo much; for 8 times 1140 is | 9120 |
| and 8 times 8 | 64 |
| In all | 9184 |
Which taken out of the remaine 9184, there now remaines nothing: which ſhews the ſquare 334084 is juſtly reſolved, and putting the laſt quotient 8 to the two former, 5 and 7, the whole will be 578, which is the true root required.
And the parts of it, if they be 3, are 500, 70, and 8. Or if but two 570 and 8. And the truth may be either way proved by adding the ſquares and twice the rectangles of the parts; for the ſum of them ſhall be equal to the whole ſquare 334084, as hath been ſhewed before.
It is ever certaine that there ſhall be as many figures, or places of figures in the root, as there are points in the ſquare, ordered as before.
And by reading this little ſeriouſly, may any one be able to find the root of any other ſquare whatſoever: the firſt operation in all being to take the greateſt ſquare contained under that point next the left hand, and put the root thereof for the firſt figure in a quotient.
The Secondary operation, muſt be repeated ſo often as there are remaining points; as hath been plainly ſhewed in the fore-going Example.
I will therefore adde no more, but give you two other ſquares, and their roots, leaving the Reader to extract them himſelf.
So, if there were given the ſquare 88̇30̇
The root of it according to the former practice may be found to be 94.
Or let there be given the ſquare 1̇52̇27̇56̇
The root of that will be 1234.