and Examples of Equations of the Second Degree. 4 s = √2401 = ± 7, or = ±7√−1; x= 4, or = ±3 √−1; = ± 3, or = ±4√−1. y= x y (x3 — y3) — 2 x2 y2 (x − y) + (x − y)2=157. Ans. x=4, or=-3, or =( 1±√−51); y=3, or 4, or =(-1-51). or x=±†√(−1572±2√(624+1574)) — 781, y=±{v(−157±2√(624+1574))+78}. 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. xy+6xy+8y3+(z−y)(y−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+8 y2 = 0, q2 y3 — 6 q y3 +8y3+(qy—2 y)(y2—5y+4)=0. Both these equations are satisfied by the value of y, But if we divide the first of these equations by y2, and the second by y, we have q2 y2 — 6 q y2+8 y2 + (q—2) (y2 —5y+4)=0; Examples of Substitution of Unknown Quantities. the first of which gives The value of q, q = 2, or = 4. q = 2, 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. Solution. 13 times the first equation, diminished by the 12 q5 y5 + 13 q3 y5 — 4 q y5 — 0. Which is satisfied by the value of y, and this value of y, being substituted in the given equations, produces x5 = 5, x5 = 65; which are evident impossibilities, and therefore the value y=0 is impossible. Examples of Substitution of Unknown Quantities. Dividing, then, by y5, we have 12 q5 13 93-4q=0; Now the first of the given equations becomes, by the substitution of x= qy, q5 y5+q3 y55; hence, by the substitution of the above values of q, we have (y2—y)2+(3 x y+2y)2—9 x2(2y+3)—12 y (x+2y)=0. and y = 1. or x=(-5±√-5,) and y = 1±√-5; or x=-5, 4. Solve the two equations Examples of Substitution of Unknown Quantities. 5. What two numbers are they, twice the sum of whose squares is 5 times their product, and the sum of whose sixth powers is 65. Ans. 2 and 1, or -2 and -1. 6. What two numbers are they, the difference of whose fourth powers is 65, and the square of the sum of whose squares is 169. Ans. ±2, and ± 3. 16* |