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 A r2 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+Bx+ 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 A2x2 + 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: M A2 x2+ABx+AM Ax+ B. Root. 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, A x + 1 B is the root of the square which differs from it by this remainder, that is, A2x2+A B x + AN=(A x + {B)2 + A M — { B2=0; or, transposing AM-B2, we have (A x + } B)2 = { B2 — A M. Now the square root of this last equation is Ax+B=± √ († B2 — A M), 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 x= -B+✓ (B2-4 AM) 2 A which is the same as A2 x2 + A B x + † B2 = † B2 — A M, is obtained immediately from the equation A2 x2+ A B+ A M=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 B24 AM a negative quantity, that is, B2 < 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 h-M m = A a2+Ba, M=2A a+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+A h)h = A a2+Ba+(M+Ah) h, and if the root of the equation A x2 + B x = M - is called the quadratic root of - M, and M the 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 Examples of Equations of the Second Degree. tiplication is Ah instead of h, and the division into periods is useless. 233. EXAMPLES. 1. Solve the equation 3x2+5x= 1491. Solution. The First Method of art. 230. Multiply by 3 and the product is 9 x2+15x=4473. The square completed by the addition of ( 15 )2 = (§)2 = 25 = 6·25 The Second Method of art. 232. In the second column of this form, the number at the top of the column is the root, the numbers above each line are the successive divisors, and the numbers below are the increased divisors before multiplication; and it is to be observed, that by the repetition of the increment the next divisor is obtained. We have, then, for the first root |