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multiplied by the other part, together with the square
of the other part.


Accordingly, if we multiply 10 by 2 and divide 69 by
this product we get a clue to the other part. Dividing
69 by 2 x 10, or 20, we see that the quotient is a little
greater than 3; if, then, after taking 3 times 20 from 69
there is left the square of 3, we have the root. Plainly
this is the case :

169/10 +3

20 69


9= 3
Now, this work might be written somewhat more neatly,
thus :

16910 +3

+3 69


It may be further simplified by leaving out unneces-
sary zeros, thus:





The pupil is now in a position to find the square
root of all numbers expressed by three or four digits.
It would be well, before considering the square root of

larger numbers, to examine for the square root of such
numbers as 1.69, 27.09. The pupil will see at once
that the square root of the former number lies between
1 and 2, that of the latter between 5 and 6, and can
easily be led to complete the process of extracting the
roots, finding as results 1.3 and 5.3. He will thus dis-
cover for himself that the problem is not different from
the one already solved.

We are now ready to examine for the square root
of larger numbers. Write first the following table:

100% =
2009 =
300o =


400* = 160000
500' = 250000
600o = 360000
700' = 490000
800° = 640000
900' = 810000

[1000ʻ = 1000000]

A study of the table will lead to the conclusion that
the square roots of numbers expressed by 5 or 6 digits
are numbers expressed by 3 digits, and that, if expressed
by 5 digits, the first (to the left) digit of the root is
determined by the first (to the left) figure of the num-
ber, and if expressed by 6 digits, by the first two digits
of the number. Thus, the square root of 16900 will
have 1 as the hundreds digit, while that of 270900 will
have 5 as the hundreds digit. Next, by multiplication,
we find that

230* = 52900

240' = 57600

Consequently the square root of (say) 54756 must lie be-
tween 230 and 240—that is, must have 23 as its first two
digits. The first three digits of the number 54756 are
sufficient to show that this must be the case. Now, sup-
pose we seek the square root of 54756.
Plainly, the first part of the root is 200:

54756 200

Now, had we been seeking the square root of 52900,
which is 230—that is, consists of two parts, 200 and
30—we should have worked thus :

[blocks in formation]

Then plainly we see how, in finding the square root of
54756, to determine the second figure:

[blocks in formation]

+ 30}

We have yet to find the units digit of the root. But at this point we may say that the root consists of two parts, one 230, and the other to be found, and may proceed as in the earlier case :

54756/200 + 30+4

400 +30

230 X 2= 460





The work may now be shortened :


43 147

464 1856


After the pupil has been exercised in extracting the roots of numbers expressed by 5 or 6 digits, he will find no difficulty in determining the square roots of such numbers as 547-56, 5-4756, :054756. The extension to numbers expressed by a higher number of digits will be easy, and the need for marking off into periods of two, starting from the decimal point, as well as its full significance, will have been realized by the pupil.

Up to this point we have spoken of numbers whose

square root can be extracted; it will be next in order to deal with the approximations to square roots—for example, the square root of 2, 5, etc.; but as this involves nothing essentially new it will not be here discussed.

We shall conclude this part of the work by calling attention to the extraction of the square root of a fraction. Since

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In the case of fractions whose denominators are numbers whose roots can not be exactly determined, we should proceed as follows:

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an artifice the value of which is apparent.

Cube Root.—The method of teaching square root has been presented in such detail that very few words will suffice on the subject of cube root.

From examples such as 3 = 27 the meaning of cube and cube root will be brought out, and use may be made of the geometrical illustration of the cube. The pupil should commit to memory the following table :

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