Squaring both members of (1) and (2), Adding (3) and (4), and omitting the common fac Now, if a – b is a perfect square, its root may be represented by c. Substituting in (V) and (8), and extracting the square root of each member of both equations, (axiom 5), we have, These values, substituted in (1) and (2), give, The square root of the given quantities may be extracted when a? b is a perfect square, and the roots may be obtained, by substitution, from (9) and (10). EXAMPLES. 1. Required the square root of 14 +675=14+V180. Here, a=14, b= 180, and c= V196 – 180 = 4: hence, V 14 + 6 V5 = 14 + 4 + 2 14 - 4 2 3+V5. 2. Required the square root of 18 – 2V77. Here, a = 18, b=308, and c= V324 — 308 = 4; hence, 3. Required the square root of 94 + 42V5. Ans. y + 375. 4. Required the square root of 28 + 10V3. Ans. 5 + 13. 6o. Operations on Imaginary Quantities. 136. An imaginary quantity has been defined to be an indicated even root of a negative quantity. The rule deduced for multiplying radicals requires some modification, when applied to imaginary quantities. By the rule already deduced, the product of v-4 by ✓ - 3 would be equal to V12; whereas, the true product is -V12, as will be shown hereafter. Every imaginary quantity of the second degree can, by principle 1°, (Art. 125), be resolved into two factors, one of which is ✓ – 1; the other factor may be either rational or irrational. Thus, 7–4=2V-1, V-3=V3x V-1, varav -1. The factor, V-1, is called the imaginary factor, and the other one is called its coefficient. Thus, in the expression, V3 V-1, the factor V3 is the coefficient of the imaginary factor ✓– 1. When several imaginary factors are to be multiplied together, we first reduce them to the form, av – 1. We can then multiply together the coefficients of the imaginary factor by known rules. It remains to deduce a rule for multiplying together the imaginary factors, or what is the same thing, for raising the imaginary factor to a power whose exponent equal to the number of factors. The first power of V - 1, is ✓-1; the second power, by the definition of square root, is – 1; the third power, is the product of the first and second powers, or -1xV1x V-1 v=1; the fourth power, is the square of the second power, or +1; the fifth, is the product of the first and fourth, that is, it is the same as the first; the sixth, is the same as the second; the seventh, the same as the third ; the eighth, the same as the fourth; the ninth, again, the same as the first; and so on indefinitely, as shown in the table, n being any whole number. To show the use of this table, let it be required to find the continued product of — 4, V – 3, V – 2, V 7, and 8. Reducing these expressions to the proper form, and indicating the multiplication, we have, 2v1 x V3V-1 xV2V-1xViV-1x2v2V 1. Changing the order of the factors, (2 X V3 X V2 X V1 x 2V2) (V — 1)5. Hence, the product is equal to, 8V21 x V-1=8V — 21. EXAMPLES. Perform the multiplications indicated below: 1. Vx V = 62. Ans. a x b(V-1)2 = – ab. 2. V-a? x V — 62 x v. Ans. abc (V— 1), = - abcv — 1. 3. V = a x V = 72 x v CZ x V = d2. Ans. abcd (V–1)4 = abcd. 4. (4 + V— 2) (3 – V–2). Ans. 14 - V–2. 5. (2 – V— 2) (2 – V–2). Ans. 2 – 4V — 2. 6. (3 – V — 2) (3 + V – 2). Ans. 11. From what precedes, it follows that the only radical parts of any power of an expression of the form, a Łov-1, will be of the form CV - 1. Properties of Imaginary Quantities. 138. 1o. A quantity of the form, av. – 1, cannot be equal to the sum of a rational quantity and a quantity of the form, bv1 For, if so, let us have the equality, squaring both members, we have, a? = x2 + 26xv=1-62; transposing, and dividing by 26x, 72 - a2 - 22 V-1 26x an equation which is manifestly absurd, for the first member is imaginary, and the second real, and no imaginary quantity can be equal to a real quantity; hence, the hypothesis is absurd; and, consequently, the principle enunciated is true. In the same way, it may be shown that no radical of the second degree can be equal to an entire quantity plus a radical of the second degree. 2o. If, a +bV-1 = x + yv — 1, then a = x, and b = Y For, by transposition, we have, bV-1 = (x – a) + yv- 1; 9 |