Page:Encyclopædia Britannica, Ninth Edition, v. 1.djvu/585

Rh CUBIC EQUATIONS.] Let A L G E B K A x c~Zx~ 29 . .16 is the equation required. (3.) a*, 6, c 2. If y = # 2, the values of y are a 2 , b 2 , c 2. Accordingly we require to throw the given equation into a form which involves no odd powers of x. This is done as follows : x 3 - x&quot;- + 2jc - 3 = ^(a- 8 + 2) - (x^ + 3) = ; x(x 2 + 2) = x 2 + 3. squaring, x 2 (x 2 + 2) 2 = (x 2 + 3) 2 , or or j&amp;gt; - 2y - 9 =. a b c be ac ab Let 2/=- = - is the equation required. Ex. 2. Two roots of the equation 5 = are 1 and 5. Find the other roots. The quantity on the left hand side of the equation is (Art. 78) divisible by (x -!)(#- 5), or by x 2 - 6x + 5. The quotient is x 2 - IQx + 21, which, being put =0, gives 3 and 7, the roots required. Ex. 3. The equation x 3 - 4# 2 + x + r has one root, 3; find r and the other roots. Write 3 for x, then r = 6, and the equation may be written (a; - 3)(x 2 - x - 2) = , which gives x = 2, x -I. Ex. 4. The equation y? + a 2 - 1 Gx - 1 6 = has two roots of the form + a, -a; find them. If we write - x for x, we get the equation x 3 -3? -lGx+ 16 = 0, which has also two roots, - a, +a, .: x 2 - a 2 is a common measure of the two quantities. But x 1 - 1C is easily found to be a common measure of the two quantities,. . a = 4. Ex. 5. The roots of the equation are in arithmetical progression; find them. If a, a + b, a + 26 be the roots, their sum is 3(a + 6), i.e., three times the middle root. But (Art. 79) their sum is 6,. . a + b = 2, also a(a + b) (a + 26) = G , i.e., o(4 - a) = 3, a 2 - 4a + 3 = , a = 1, a = 3. Ex. 6. The three roots of the equation a? - 7x 2 + lGx -8 = are in geometrical progression; find them. Let a, ar, ar 2 be the roots; then their product is (a? ) 3 , . . (ar) 3 = 8, and ar = 2,. . their sum a + ar + ar 2 = 7, which gives a = 1, r = 2 , and 1 2, 4 are the roots. . SECT. XI. SOLUTION OF CUBIC EQUATIONS. 88. Cubic equations, like all equations above the first degree, are divided into two classes: they are said to be pure when they contain only one power of the unknown quantity; and adfected when they contain two or more powers of that quantity. Pure cubic equations are therefore of this form, x 3 = 125, ora 3 = -27, or, in general, y? = r; and hence it appears that a value of the simple power of the unknown quantity may always be found without difficulty, by extracting the cube root of each side of the equation; thus, from the first of the three preceding examples we find x + 5, from the second x = - 3, and from the third, x = Jr. It would seem at tirst sight that the only v alue which x can have in the cubic equation x 3 = r, or putting r = c 3 , x 3 - c 3 = 0, is this one, x = c; but since x 3 - c 3 may be re solved into these two factors, x c and x 2 + ex + c 2, it fol lows, that besides the value of x already found, which results from making the factor x - c = 0, it has yet other two values, which may be found by making the other fac tor x 2 + ex + c 2 = ; and accordingly, by resolving the qua dratic equation x 2 + ex = - c 2, we find these values to be -c + V-3c 2 _ c _V-3c 2 -1 + V^3 -1-V^3 -p- - and, or cancl c. Thus it appears, that any cubic equation of this form, x 3 = c 3, or x 3 c 3 = 0, has these three roots, X = C, X = c,x = the first of which is real, but the two last are imaginary. If, however, each of the imaginary values of x be raised to the third power, the same results will be obtained as from the real value of x; the original equation 3 -c 3 = may also be reproduced, by multiplying together the three factors x - c, x- -l+V-3 c, and x -l-V-3 89. Let us now consider such cubic equations as have all their terms, and which are therefore of this form, where A, B, and C denote known quantities, either posi tive or negative. It has been shown (Art. 84) how an equation having all its terms may be transformed into another which wants the second term ; therefore, assume x = y -, as directed o in that article ; then, by proper substitution, the above equation will be changed into another of this form, 2/ 3 + qy + r = , where q and r denote known quantities, whether positive or negative ; now the roots of this equation being found, it is evident that those of the former may be readily ^ obtained by means of the assumed equation x = y -. o Resuming, therefore, the equation y 3 + qy + r = 0, let us suppose y = v + z, and it becomes + qv + qz V =. + r ) Thus we have a new equation, which, as it involves two unknown quantities, v and z, may be resolved into any two others, which will simplify the determination of those quantities. Now, it appears, that the only way in which we can divide that equation into two others, so as to simplify the question, is the following : 3v 2 z + 3vz 2 + qv + qz, t) 3 + 2 3 + r= Oi The first of these may also be expressed thus, Hence, we must either suppose that v + z = Q, or that 3vz + q = ; but the former supposition cannot be admitted without supposing also that y = 0; therefore we must adopt the latter. So that to determine v and z we have these two equations, Bvz + q = 0, v 3 + z 3 + r =. From the first, we find vz= - |, and. . v 3 z 3 = - ~. This 3 27 reduces the second equation to a quadratic in v 3, viz. Q^ 3 - =0, the solution of which equation is