sor; and multiplying all these last terms by the c of the root, we obtain 3 a2 c + 3 a c2 + c3, a product equal to the dividend; consequently the extraction is completed, and the required root is a + c; which, as before, may be proved by raising the root ac to the 3d power, which will be found to be the same with the quantity given in the ques tion. Another method is, when the second term of the root is found to add together three times the square of the first term, three times the product of that part multiplied by the second term, and the square of this second; the sum will be the divisor for the second step, which multiplied by the second term of the root, will give the product desired. This will be illustrated by using arithmetical instead of algebraic characters; supposing a to be 8 and c to be 5, the quantity 8+ 5 will be cubed, and the cube root again extracted, in the algebraic form, as follows. Here the cube root of the first term, 512, being 8, this number becomes the first term of the root, and its cube, 512, subtracted from the given cube, leaves no remainder; then the other terms of the given cube are brought down, and a divisoris formed by taking three times the square of the first term of the root 8, or 192, by which dividing 960, the first term of the new dividend, the quotient 5 becomes the second term of the root; then taking three times the product of these two terms, which is 120, and the square of the second term, 25, we have the whole of the new divisor, which being multiplied by the second term, 5, the product is 960 + 600 + 125, equal to the dividend; consequently the extraction is finished and 8+ 5 is the cube root required. Or thus, the sum of the two terms 8 and 5 being 13, the cube of 13 is 2197, which is equal to the sum of 512, the cube of the first term 8, together with 960 and 600, the intermediate terms of the cube, and 125, the cube of the last term 5, as is here shown. In calculations where algebra is employed, it is generally the object, by means of certain known quantities, to discover others that are unknown, but which are stated to stand in given proportions or relations to those known quantities; and this is done by discovering what portion of such known quantities is equal to those required; this portion is termed the value of the unknown quantity, and the statement of such a value is termed an equation. Thus, in the equation a + m − n = x, if the values of a, m, and n be known, their sum, or x, must also be determined. An equation is said to be resolved, when the known quantities are all placed on the one side of the mark of equality, and the unknown quantities on the other side; and the value of the unknown quantities is called the root of the equation. When an equation expresses the value of the first power or an unknown quantity, it is called a simple equation; when it contains the second power, or square of the unknown quantity, it is called a quadratic equation; when it exhibits the cube or third power, it is a cubic equation; the 4th power is a biquadratic equation; and so on with respect to all higher powers. 1st. When the unknown quantity is connected with a known quantity, and their conjunct value is known, the unknown quantity may be transported to the one side of the equation, and the known quantities to the other, by what is called transposition, as in the following example, where x-5 is given equal to 20; or thus x-5= 20. Here we see, that if we subtract 5 from the value of X, the remainder will be 20, consequently 20 ÷ 5 = 25 will be the value of x, as here shown. So that any quantity may be removed from the one side of the equation to the other, that is, transposed, merely by changing the prefixed sign from +to-, or from to +. Again, let 6x+4= 9x 20, required the value of x. Here as 6x wants 4 to be equal to 9x-20, the difference 4 may be removed to the opposite side of the equation, changing its sign to -, thus giving a new equation 6.x 9x 20 4, that is 9x 24, which sum, 24, is therefore the difference between 6x and 9x, that is, 8x; and if 3x be equal to 24, x alone must be equal to the third part of 24, or equal to 8, which is the value required, as may be proved by the following statement. From this rule it follows, that when any quantity is found on both sides of an equation, it may be expunged without in the least affecting the solution: thus if a + x = m— c+a, the quantity a may be taken away from both sides of the equation, and the remainder x will be equal to the remainder mc, or, arithmetically, 6+4=7-3+6 4=7-3 2nd. When the unknown quantity is multiplied by any number or quantity, this number or quantity may be remov ed by dividing by it the other terms of the equation: thus, if 5 m be equal to 25, m will be equal to 25 divided by 5, that is, m will be equal to 5. Again, let m x be equal to a + b, required the value of m. = 12, then m or x 5 × 6 = 30 = 18 + 12 = 30, and 3rd. When any term of an equation is a fraction, it may be removed by multiplying all the other terms by the deno a minator of the fractional part : for instance, if + + 6 = 24, 3 multiplying the other terms by the denominator 3, we have a new equation, a + 18 = 72, and consequently a = 72 — 18 = 54,the third part of which = 18, added to 6, will be equal to 24, agreeably to the statement of the example. 4th. When there are several fractions in an equation, all the denominators may be removed, provided some number can be found which is a multiple of each denominator, by which to multiply all the numerators of the terms of theequation; thus, if + +, be equal to 22, we may employ 2 |