(24) }xy=√x2+y2+x+y, x2+y2=(x+y)2—{xy2. 16 Ans. x=3. Ans. t=±√√6. Ans. x=625, y=16. Ans. x=5, y=±4. Ans. x=4 or 2, y=2 or -4. (31) x3—y3=56, x—y=xy Ans. Ans. x=5 or 2√3, y=4 or 10. Ans. x 64 or 8, y=8 or 64. 18 or ±√8, y=32 or 1024. Ans. x=5 or —1, y=1 or —5. 181. We have seen that an equation of the form arb has two roots, or that there are two quantities which, when substituted for x in the original equation, will render the two members identical. In like manner, we shall find that every equation which involves r in the third power has three roots; an equation which contains has four roots; and it is a general proposition in the theory of equations that an equation has as many roots as it has dimensions. 182. The above method of solving the equation ar"=b will give us only one of the n roots of the equation if n be an odd number, and two roots if n be an even number. Such a solution must, therefore, be considered imperfect, and we must have recourse to different processes to obtain the remaining roots. This, however, is a subject which we must postpone for the present. SOLUTION OF COMPLETE QUADRATICS, CONTAINING ONE UNKNOWN QUANTITY. 183. In order to solve the general equation ar+br=c, let us begin by dividing both members by a, the coefficient of x2; the equation then becomes This form of the quadratic equation may be produced by multiplying together two simple equations. Suppose x-a-0, x-b=0; .. (x-a)(x—b)=0, which is satisfied by making r=a, or x=b. Multiplying the two factors (-a) and (x-b), the equation becomes x2—(a+b)x+ab=0. . . . (1) Substituting first a, and then b, for x, this may be written either Putting in equation (1) above p, in place of (a+b), and -q in place of ab, it assumes the form By addition and subtraction, a=- − } p + 1⁄2 √ p2 + 4q b = p- √p2+4q. As a and b are the values of r, and differ only in the sign of the radical part, both may be written together thus: Hence the following rule for resolving a complete or adfected quadratic equation. Reduce the given equation to the form x2+px-q=0 by clearing of fractions, transposing all the terms to the first member, and dividing throughout by the coefficient of the square of the unknown quantity. The equation being thus prepared, the value of the unknown quantity will be equal to the coefficient of its first power with the sign changed, the square root of the square of this coefficient -4 times the known terms of the equation. The expression x=−¦p±¦√p2+4q may, by passing the under the radical, be written x=-p√(p)2+q, which, translated into a rule, is often the more convenient form. |