also when x=a in the given equation, y = (b-c)2 in the transformed equation; COR. If a, b, c are real, (b −c)2, (c − a)2, (a - b)2 are all positive; therefore 27r2+4q3 is negative. Hence in order that the equation x3+qx+r=0 may have all its roots must be negative. real 27r2+4q3 must be negative, that is (2)2 + (3) * 3 If 27r2+4q3=0 the transformed equation has one root zero, therefore the original equation has two equal roots. If 27r2+4q3 is positive, the transformed equation has a negative root [Art. 553], therefore the original equation must have two imaginary roots, since it is only such a pair of roots which can produce a negative root in the transformed equation. integral coefficients, and unity for the coefficient of the first term. 2. Transform the equation 3x1 − 5x3+x2 −x+1=0 into another the coefficient of whose first term is unity. 7. Solve the equation 3x3 – 22x2 + 48x − 32 =0, the roots of which are in harmonical progression. 8. The roots of 23-11x2+36x-36=0 are in harmonical progression; find them. 9. If the roots of the equation x3 — ax2 + x-b=0 are in harmonical progression, shew that the mean root is 36. 10. Solve the equation 40.1 - 22x3 − 21x2+2x+1=0, the roots of which are in harmonical progression. Remove the second term from the equations: 3 2 exceed by the corresponding roots of the given equation. 16. Diminish by 3 the roots of the equation 25-4x4+3x2-4x+6=0. 17. Find the equation each of whose roots is greater by unity than a root of the equation x3 – 5x2+6x-3=0. 18. Find the equation whose roots are the squares of the roots of x2+x3 + 2x2+x+1=0. 19. Form the equation whose roots are the cubes of the roots of x3+3x2+2=0. If a, b, c are the roots of 3+qx+r=0, form the equation whose 27. Shew that the cubes of the roots of x3+ax2+bx+ab=0 are given by the equation a3+a3x2+b3x+a3b3=0. 28. Solve the equation -5x1- 5x3+25x2+4x-20=0, whose roots are of the form a, a, b, −b, c. 29. If the roots of x3+3px2+3qx+r=0 are in harmonical progression, shew that 2q3=r (3pq − r). CUBIC EQUATIONS. 575. The general type of a cubic equation is x2 + Px2 + Qx+ R = 0, but as explained in Art. 573 this equation can be reduced to the simpler form x3 + qx + r = 0, which we shall take as the standard form of a cubic equation. 576. To solve the equation a3 + qx + r = 0. Let x=2 x3 = y3 + z3 + 3yz (y + z) = y3 + z3 + 3yzx, and the given equation becomes y3 + z3 + (3yz +q) x + r = 0. At present y, z are any two quantities subject to the condition that their sum is equal to one of the roots of the given equation; if we further suppose that they satisfy the equation 3yz+q=0, they are completely determinate. We thus obtain we obtain the value of x from the relation x = y + z; thus (1), (2), The above solution is generally known as Cardan's Solution, as it was first published by him in the Ars Magna, in 1545. Cardan obtained the solution from Tartaglia; but the solution of the cubic seems to have been due originally to Scipio Ferreo, about 1505. An interesting historical note on this subject will be found at the end of Burnside and Panton's Theory of Equations. 577. By Art. 110, each of the quantities on the right-hand side of equations (1) and (2) of the preceding article has three cube roots, hence it would appear that x has nine values; this, however, is not the case. For since yz=-2, the cube roots are 3' to be taken in pairs so that the product of each pair is rational. Hence if y, z denote the values of any pair of cube roots which fulfil this condition, the only other admissible pairs will be wy, w2z and w3y, wz, where w, w2 are the imaginary cube roots of unity. Hence the roots of the equation are 578. To explain the reason why we apparently obtain nine values for x in Art. 576, we observe that y and z are to be found -; but in the process of from the equations y3 +≈3+r=0, yz 3 solution the second of these was ohanged into y3z3: H. H. A. which 31 values of x are solutions of the cubics x23 + wqx + r = 0, x2+w3qx+r= 0. 579. We proceed to consider more fully the roots of the equation x3 + qx + r = 0. 3 2.2 (i) If + is positive, then y3 and 3 are both real; let 4 27 y and z represent their arithmetical cube roots, then the roots y + 2, wy + w3z, w3y + wz. are The first of these is real, and by substituting for w and w3 the other two become (ii) If + is zero, then y3=23; in this case y = 2, and 4 27 the roots become 2y, y(w+w2), y(w+w2), or 2y, −y, −y. 93 + is negative, then 4 27 y3 and 23 are imaginary ex (iii) If pressions of the form a + ib and a ib. Suppose that the cube roots of these quantities are m + in and m − in; then the roots of the cubic become A (m + in) w + (m − in) w3, or —m-n √3; (m + in) w2 + (min) w, or m+n√3; which are all real quantities. As however there is no general arithmetical or algebraical method of finding the exact value of the cube root of imaginary quantities [Compare Art. 89], the solution obtained in Art. 576 is of little practical use when the roots of the cubic are all real and unequal. This case is sometimes called the Irreducible Case of Cardan's solution! 580. In the irreducible case just mentioned the solution may be completed by Trigonometry as follows. Let the solution be x = (a + ib)3 + (a — ib)3'; |