Cases of imaginary Solution. 29. Some merchants engage in business; each contributes a thousand times as many dollars as there are partners. They gain in this business $ 2560; and it is found that this gain is exactly half their own number per cent. How many merchants are there? Ans. 8. 30. Find three numbers such, that the square of the first multiplied by the second is 112; the square of the second multiplied by the third is 588; and the square of the third multiplied by the first is 576. Ans. 4, 7, 12. 227. Corollary. When the solution of a problem gives for either of its unknown quantities only imaginary values, the problem must be impossible. 228. EXAMPLE. In what case would the value of the unknown quantity in example 13 of art. 226 be imaginary? and why should the problem in this case be impossible ? that is, when the product of the sum and difference required to be greater than the square of a. Now if the required number is z, this product is (a + x) (a - x) = a2 and, therefore, less than a2 x2; Equations of the Second Degree. CHAPTER VI. EQUATIONS OF THE SECOND DEGREE. 229. It may easily be shown, as in art. 120, that any equation of the second degree with one unknown quantity, may be reduced to the form A x2 + B x + M = 0, in which Aro denotes the aggregate of all the terms multiplied by the second power of the unknown quantity, Br denotes all the terms multiplied by the unknown quantity itself, and M denotes all the terms which do not contain the unknown quantity. 230. Problem. To solve an equation of the second degree with one unknown quantity. Solution. Having reduced the given equation to the form A x2 + B x + M = 0, we could easily reduce it to an equation of the first degree, by extracting its square root, if the first member were a perfect square. But this cannot be the case, unless the first term is a perfect square; the equation can, however, always be brought to a form in which its first term is a perfect square, by multiplying it by some quantity which will render the coefficient of the first term a perfect square, multiplying by this coefficient itself, for instance; thus the given equation multiplied by A becomes Ar2 + A B x + A M = 0. Equations of the Second Degree. Now that the equation is in this form, we can readily ascertain whether its first member is a perfect square, by attempting to extract its root, as follows: A2 x2 + A B x + AMA x + B. Root. ABxAM2 Az ABx+B2 AM-B2 Rem. so that the first member is a perfect square only when the remainder is zero, that is, AM - B2 = 0 ; and, in every other case, Ax+B is the root of the square which differs from it by this remainder, that is, A2x2+ABx+AN = (Ax+B)2 + AM-B2= 0; or, transposing AM-1 B2, we have (Ax + B)2 = ¦ Bo— A M. Now the square root of this last equation is Ax + B = ± √ (¦ B2 — AM), which, solved as an equation of the first degree, gives in which either of the two signs + or -, may be used of the double sign ±, and we thus have the two roots of the given equation -B+√(B2-4 AM) A2 x2 + A B x + + B2 = + B2 — А М, is obtained immediately from the equation A2 x2 + A B x + AM = 0, by transposing AM to the second member, and adding B2 to both members. Hence To solve an equation of the second degree with one unknown quantity. Reduce it as in arts. 112 and 118, transposing all the terms which contain the unknown quantity to the first member, and the other terms to the second member. Multiply the equation by any quantity, (the least is to be preferred,) which will render the coefficient of the second power of the unknown quantity an exact square. Add to this equation the square of the quotient, arising from the division of the coefficient of the first power of its unknown quantity, by twice the square root of the coefficient of the second power of its unknown quantity. Extract the square root of the equation thus augmented, and the result is an equation of the first degree, to be solved as in art. 121. Affected Quadratic Equation. 231. Corollary. When we have B2-4 AM a negative quantity, that is, Bo < 4 A M, the roots of the given equation are imaginary. 232. Scholium. The preceding method of solving quadratic equations gives the form of the roots in all cases, but otherwise it has no advantage in the solution of a numerical equation over the solution given in Chapter IV. The method of art. 182, applied to this case, gives m = A a2 + Ba, M=2Aa + B. But the process may be abbreviated precisely as in the case of the square root in art. 189, by observing that A(a+h)2+B(a+h) = A a2+Ba+(2Aa+B+Ah)h = A a2+Ba+(M+Ah)h, and if the root of the equation A x2 + B x = - M M the is called the quadratic root of - M, and quadratic power of its root, the rule for extracting its root is the same as that for extracting the square root in art. 189, except that QUADRATIC must be substituted for SQUARE, the divisor is, in each case, 2 Aa + B instead of 2 a, the addition to the divisor before mul |