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Therefore, when the two members of an inequality are multiplied or divided by a number expressed algebraically, it is necessary to ascertain whether the multiplier or divisor is negative; for, in that case, the inequality would exist in a contrary sense.

4. It is not permitted to change the signs of the two members of an inequality unless we establish the resulting inequality in a contrary for this transformation is evidently the same as multiplying the two members by −1.

sense;

5. Both members of an inequality between positive numbers can be squared, and the inequality will exist in the same sense.

Thus from 5>3, we deduce 25>9; from a+b>c, we find (a+b)2>c2.

6. When both members of the inequality are not positive, we cannot tell before the operation is performed, in which sense the resulting inequality will exist.

For example, -2<3 gives (-2) or 4<9; but -3>-5 gives, on the contrary, (-3) or 9<(-5)2 or 25.

We must then, before squaring, ascertain whether the two members can be considered as positive numbers.

EXAMPLES.

1. Find the limit of the value of x in the expression

52–6>19.

2. Find the limit of the value of x in the expression

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3. Find the limit of the value of x in the expression

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4. Find the limit of the value of x in the inequalities

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5. The double of a number diminished by 5 is greater than 25, and triple the number diminished by 7, is less than double the number increased by 13. Required a number which shall satisfy the conditions.

By the question, we have

2x-5>25.

3x-72x+13.

Resolving these inequalities, we have x>15 and 20. Any number, therefore, either entire or fractional, comprised between 15 and 20, will satisfy the conditions.

6. A boy being asked how many apples he had in his basket, replied, that the sum of 3 times the number plus half the number, diminished by 5 is greater than 16; and twice the number diminished by one third of the number, plus 2 is less than 22. Required the number which he had.

Ans. 7, 8, 9, 10, or 11.

CHAPTER III.

Extraction of the Square Root of Numbers. Formatron of the Square and Extraction of the Square Root of Algebraic Quantities. Calculus of Radicals of the Second Degree. Equations of the Second Degree.

116. The square or second power of a number, is the product which arises from multiplying that number by itself once: for example, 49 is the square of 7, and 144 is the square of 12.

The square root of a number is a second number of such a valuc, that, when multiplied by itself once the product is equal to the given number. Thus, 7 is the square root of 49, and 12 the square root of 144: for 7x7=49, and 12×12=144.

The square of a number, either entire or fractional, is easily found, being always obtained by multiplying this number by itself once. The extraction of the square root of a number, is however, attended with some difficulty, and requires particular explanation. The first ten numbers are,

1, 2, 3, 4, 5, 6, 7,

and their squares,

1, 4, 9, 16, 25, 36, 49, and reciprocally, the numbers of the first line of the corresponding numbers of the second.

8, 9, 10,

64, 81, 100.

are the square roots We may also remark

that, the square of a number expressed by a single figure, will contain no figure of a higher denomination than tens.

The numbers of the last line 1, 4, 9, 16, &c., and all other numbers which can be produced by the multiplication of a number by itself, are called perfect squares.

It is obvious, that there are but nine perfect squares among all the numbers which can be expressed by one or two figures: the square roots of all other numbers expressed by one or two figures will be found between two whole numbers differing from each other by unity. Thus, 55 which is comprised between 49 and 64, has for its square root a number between 7 and 8. Also, 91 which is comprised between 81 and 100, has for its square root a number between 9 and 10.

Every number may be regarded as made up of a certain number of tens and a certain number of units. Thus 64 is made up of 6 tens and 4 units, and may be expressed under the form 60+4=64. Now, if we represent the tens by a and the units by b, we shall a+b= 64

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Which proves that the square of a number composed of tens and units contains, the square of the tens plus twice the product of the tens by the units, plus the square of the units.

117. If now, we make the units 1, 2, 3, 4, &c., tens, by annexing to each figure a cipher, we shall have,

10, 20, 30, 40, 50, 60,

and for their squares,

70, 80, 90, 100

100, 400, 900, 1600, 2500, 3600, 4900, 6400, 8100, 10000. from which we see that the square of one ten is 100, the square of two tens 400; and generally, that the square of tens will contain no figure of a less denomination than hundreds, nor of a higher name than thousands.

60.84

EXAMPLE I.-To extract the square root of 6084. Since this number is composed of more than two places of figures its roots will contain more than one. But since it is less than 10000, which is the square of 100, the root will contain but two figures: that is, units and tens. Now, the square of the tens must be found in the two left hand figures which we will separate from the other two by a point. These parts, of two figures each, are called periods. The part 60 is comprised between the two squares 49 and 64, of which the roots are 7 and 8: hence, 7 is the figure of the tens sought; and the re. quired root is composed of 7 tens and a certain number of units. The figure 7 being found, we write it on the right of the given number, from which we separate it by a vertical line: then we subtract its square 49 from 60, which leaves a remainder of 11, to which we bring down the two next figures 84. The result of this operation 1184, contains twice the product of the tens by the units plus the square of the units. But since tens multiplied by units cannot give a product of a less name than tens, it fol

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49

7X2=14.8 | 118.4

118 4

lows that the last figure 4 can form no part of the double product of the tens by the units: this double product is therefore found in the part 118, which we separate from the units' place 4 by a point.

Now if we double the tens, which gives 14, and then divide 118 by 14, the quotient 8 is the figure of the units, or a figure greater than the units. This quotient figure can never be too small, since the part 118 will be at least equal to twice the product of the tens by the units: but it may be too large; for the 118 besides the double product of the tens by the units, may likewise contain tens arising from the square of the units. To ascertain if the quotient S expresses the units, we write the 8 to the right of the 14, which gives 148, and then we multiply 148 by 8. Thus, we evidently form, 1st, the square of the units: and 2d, the double product of the tens by the units. This multiplication being effected, gives for a product 1184, a number equal to the result of the first operation. Having subtracted the product, we find the remainder equal to 0: hence 78 is the root required.

Indeed, in the operations, we have merely subtracted from the given number 6084, 1st, the square of 7 tens or 70; 2d, twice the product of 70 by 8; and 3d, the square of 8: that is, the three parts which enter into the composition of the square of 70+8 or 78; and since the result of the subtraction is 0, it follows that 78 is the square root of 6084.

8.41 29

4

2X2 4.9 44.1

Ex. 2. To extract the square root of 841. We first separate the number into periods, as in the last example. In the second period, which contains the square of the tens, there is but one figure. The greatest square contained in 8 is 4, the root of which is 2: hence 2 is the fi. gure of the tens in the required root. Subtracting its square 4 from 8, and bringing down 41, we obtain for a result 441.

44 1

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