root sought. In the same manner may the roots of the following quantities be found. 1. What is the 2d or square root of 16a2b1c ? 59. CASE II. When the given quantity is compound, and it is required to extract its square root. RULE. Arrange the terms according to the powers of the same letter, so that the highest power may be first, and the next highest in the second term, and so on. Extract the square root of the first term by Case I., and place its root for the first term of the root. Subtract its square from the first term, and there will be no remainder. Bring down the next two terms for a dividend, and for a divisor take twice the part of the root already found. Divide the first term of the dividend by the divisor, and place the quotient both in the root and in the divisor. Multiply the divisor thus completed by the term last placed in the root, and subtract the product from the dividend; to the remainder, if any, bring down the next two terms, and proceed thus till the root is found. The above rule will be obvious, by observing that the square of a+x is a2+2ax+x2, and that consequently the square root of a2+2ax+x2 will be a+x; but after we have subtracted the square of a, the remainder is 2ax+x2 =(2a+x)x, the first term of which remainder, if divided by 2a, will give the quotient x; and a being added to 2a, and then the sum multiplied by x, will leave no remainder. EXAMPLE. Extract the square root of a2+2ab+b2+ 2ac+2bc+c2. a2+2ab+b2+2ac+2bc+c2 | a+b+c root. a2 Divisor 2a+b 2ab+b2 2ab+b2 2a+2b+c 2ac+26c+c2 2ac+2bc+c2 Here, after (a+b) has been obtained in the root, it is evident that the remainder can be written 2(a+b)c+c2= {2(a+b)+cc; where a+b takes the place of a in the remainder, and is of the same form as (2a+x)x. 1. Extract the square root of 9a2+бab+b2. Ans. 3a+b. 2. Extract the square root of a2+8ax+16x2. Ans. a+4x. 3. Extract the square root of a1c2+4a2cx+4x2. Ans. a2c+2x. 4. Extract the square root of 4a2+12ab+962+16ac+ 24bc+16c2. Ans. 2a+3b+4c. 5. Extract the square root of x+4x2+2x1+9x2-4x+4. Ans. x+2x2-x+2. 60. CASE III. When the given quantity is compound, and it is required to extract its cube root. RULE. Arrange the terms as in CASE II. Find the cube root of the first term, which will be the first term of the root. Subtract its cube from the first term, which will leave no remainder. Bring down the next term, and divide it by three times the square of the root already found; the quotient will be the second term of the root. Raise these two terms to the third power, and subtract the result from the given quantity; if there be a remainder, divide its first term by three times the square of the first part of the root as before, and thus proceed till the work is finished. The third power of (a+x) is a3+3a2x+3ax2+x3; hence it is evident, that the cube root of a3+3a2x+3ax2 +x3 is a+x; taking away the cube of a, the first term of the root, the remainder is 3a2x+3ax2+x3; the first term of which divided by 3a2, gives the quotient x. EXAMPLE. Extract the cube root of a6+6ax+12a2x2 +8x3. a+6a1x+12a2x2+8x3, | a2+2x. Root. αδ. The first term of the remainder is 6a4x, which divided by 3a, which is three times the square of a2, gives 2x for quotient; and a2+2x raised to the third power, gives the quantity whose root was to be extracted, and no remainder; so that a2+2x is the root sought. 1. Extract the cube root of 27a3-27 a2x+9ax2-x3. Ans. 3a-x. 2. Extract the cube root of a +бa1b2+12a2b1+866. Ans. a2+262. 3 3. Extract the cube root of 39x2y+27xy2-27y3. Ans. x-3y. 61. Any root whatever may be extracted by the following formula: let n be the name of the root, which will be 2 for the square root, 3 for the cube root, and so on; then having arranged the terms as in the square and cube roots, extract the root of the first term by Case I., and let a represent that root, then the second term divided by nan-1, will give the second term of the root; the first and second terms of the root being raised to the nth power, and subtracted from the given quantity, the remainder, if there be any, will be such, that its first term divided by the same divisor will give the third term. The exercises in the square and cube root may be wrought by this rule. 62. Definitions. EQUATIONS. 1. An Equation is a proposition which declares the equality of two quantities expressed algebraically. This is done by writing the two quantities, one before and the other after the sign(): thus 4+x=3x-4 is an equation, which asserts the equality of 4+ and 3x-4. 2. A Simple Equation is one which, being reduced to its simplest form, contains only the first power of the unknown quantity. 3. A Quadratic Equation is one which, being reduced to its simplest form, contains the square of the unknown quantity. 4. When an Equation, after being reduced to its simplest form, contains the third power of the unknown quantity, it is called a Cubic Equation. 5. A Pure Quadratic is one into which only the square of the unknown quantity enters. 6. An Adfected Quadratic is one which contains both the first and second powers of the unknown quantity. 7. The Resolution of Equations is the determining from some given quantities the value or values of those that are unknown. The resolution of equations is effected by the application of one or more of the following axioms:— AXIOMS. (1.) If equal quantities be added to equal quantities, the sums will be equal. (2.) If equal quantities be taken from equal quantities, the remainders will be equal. (3.) If equal quantities be multiplied by the same or equal quantities, the products will be equal. (4.) If equal quantities be divided by the same or equal quantities, the quotients will be equal. (5.) If the same quantity be added to and subtracted from another, the value of the latter will not be altered. (6.) If a quantity be both multiplied and divided by the same quantity, its value will not be altered. (7.) The same powers and roots of equal quantities are equal. From the above axioms the following rules for the resolution of equations can be derived : 63. RULE 1. Any quantity can be taken from one side of an equation to the other, by changing its sign. This rule is derived from axioms 1st and 2d, as will appear evident from the following example: Let 3x-4=2x+6; if 4 be added to both sides the equality will still exist by axiom first, but the equation will become 3x=2x+6+4; where the 4 has been taken to the other side and its sign changed; so that taking a minus quantity from one side to the other, and changing its sign, is equivalent to adding that quantity to both sides; if now 2x be taken from both sides, the equation will become 3x-2x-10, where by taking 2x from both sides of the equation, it has disappeared from the second side, but has reappeared on the first; hence, taking a plus quantity from one side, and placing it on the other, with its sign changed, is equivalent to subtracting equals from equals, and it has just been shown, that taking a minus quantity from one side to the other, and making it plus, is equivalent to adding equals to equals. The solution of the above equation will now stand as under: 3x-4=2x+6, the given equation. 3x=2x+10, by transposing-4 and adding 4 and 6. 3x-2x-10, by transposing 2x. x=10, by performing the subtraction on the first side. 64. RULE 2. If, after all the unknown quantities are transposed to the first side, and the known ones to the second, the unknown quantity have a coefficient, it may be taken away by dividing both sides of the equation by it. This rule is evidently the same as axiom 4th. EXAMPLE. Given 4x+27=48-3x, to find the value of x. 4x+27=48-3x, given equation. 4x+3x=48—27, by transposing —3x and 27. .. x=3, by applying the rule. 65. RULE 3. If there are fractions in any of the terms, they may be taken away by multiplying all the terms by each of the denominators in succession; or by multiplying all the terms at once, by the least common multiple of all the denominators. It is evident that this rule is merely an application of axiom 3d, and points out when that axiom may be applied. EXAMPLE. +++4=2x, to find the value of x, x+7+7+4=2x, the given equation. 4x+2x+x+16=8x, by multiplying both sides by 4. 7x+16=8x, by collecting the like terms. .. 16=x, by subtracting 7x from both sides. 66. RULE 4. If the value of any root of the unknown quantity can be found from the equation, raise both sides to the power denoted by the root, and the value of the unknown quantity will be found. This is evident from axiom 7th. EXAMPLE. Given x2+x=+4; to find the value of x, x2+x=2+4, the given equation. 2x+x=x+8, by multiplying by 2. Rule 3d. 2x=8, by taking a from both sides. x=4, by dividing by 2. Rule 2d. ..x=16, by the Rule. 67. RULE 5. If, after the equation has been reduced to its simplest form by the preceding rules, the value of some power of the unknown quantity is obtained, its value may be found by extracting the corresponding root of both sides. This is also evident from axiom 7th. x2+3x EXAMPLE. Given =x+12; to find the value of x. x2+3x x+12, the given equation. 3 x2+3x=3x+36, by multiplying by 3. Rule 3d. .. x=6, by extracting the square root. The previous rules will be found sufficient for the solution of equations containing only one unknown quantity; the following solutions are added as examples of their application. |