x y (x3 — y3) — 2 x2 y2 (x − y) + (x — y)2 = 157. Ans. x4, or 3, or =( 1±√-51); = y=3, or 4, or =(-1±√−51). orx=±{v(−1572±2√(624+1574)) — 781, y=±{√(−1572±2√(624+1574))+781. 6. What two numbers are they, whose difference is 1, and the difference of whose third powers is 7? Ans. 1 and 2, or -2 and - 1. 7. What two numbers are they, whose difference is 3, and the sum of whose fourth powers is 257? or (±√(−79)+3) and } (±√(−79)—3). Examples of Equations of the Second Degree. 242. When the first member of one of the equations, reduced as in art. 118, is homogeneous in regard to two unknown quantities, the solution is often simplified by substituting for the two unknown quantities, two other unknown quantities, one of which is their quotient. The same method of simplification can also be employed when such a homogeneous equation is readily obtained from the given equations. 243. EXAMPLES. 1. Solve the two equations x2-6xy+8 y2 = 0, x2y +6 x y2+8 y3 + (x−2y) (y2—5y+4=0. Solution. Retaining the unknown quantity y, introduce instead of x, the unknown quantity q, such that from which the given equations become q2 y2-6 q y2+8y2 = 0, y3 q2 y3 — 6 q y3 +8 y3+(qy—2y) (y2—5y+4)=0. Both these equations are satisfied by the value of y, whence y = 0, x= qy = 0. But if we divide the first of these equations by y2, and the q2y6qy2+8 y2 + (q − 2) (y2 —5y+4)=0; being substituted in the other equation, reduces the first member to zero, and therefore y is indeterminate; that is, x and y may have any values whatever, with the limitation that x is the double of y. being substituted in the other equation, gives 2(y2-5y+4)= 0; and this value of y, being substituted in the given equations produces x5 = 5, which are evident impossibilities, and therefore the value y=0 is impossible. |
