By Leonard Eugene Dickson
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Extra info for First course in the theory of equations
Solve x4 − 3x2 + 6x − 2 = 0. 4. Solve x4 − 2x2 − 8x − 3 = 0. 5. Solve x4 − 10x2 − 20x − 16 = 0. 49. Roots of the Resolvent Cubic Equation. Let y1 be the root y which was employed in §48. Let x1 and x2 be the roots of the first quadratic equation (18), and x3 and x4 the roots of the second. Then x1 x2 = 21 y1 − n, x3 x4 = 12 y1 + n, x1 x2 + x3 x4 = y1 . If, instead of y1 , another root y2 or y3 of the resolvent cubic (17) had been employed in §48, quadratic equations different from (18) would have been obtained, such, however, that their four roots are x1 , x2 , x3 , x4 , paired in a new manner.
Solve y 3 − 15y + 4 = 0. 2. Solve y 3 − 2y − 1 = 0. 3. Solve y 3 − 7y + 7 = 0. 4. Solve x3 + 3x2 − 2x − 5 = 0. 5. Solve x3 + x2 − 2x − 1 = 0. 6. Solve x3 + 4x2 − 7 = 0. 47. Trigonometric Solution of a Cubic Equation with ∆ > 0. , if R is negative, they can be computed simultaneously by means of a table of cosines with much less labor than required by Cardan’s formulas. To this end we write the trigonometric identity cos 3A = 4 cos3 A − 3 cos A in the form z 3 − 43 z − 41 cos 3A = 0 (z = cos A).
3. Solve x4 − 3x2 + 6x − 2 = 0. 4. Solve x4 − 2x2 − 8x − 3 = 0. 5. Solve x4 − 10x2 − 20x − 16 = 0. 49. Roots of the Resolvent Cubic Equation. Let y1 be the root y which was employed in §48. Let x1 and x2 be the roots of the first quadratic equation (18), and x3 and x4 the roots of the second. Then x1 x2 = 21 y1 − n, x3 x4 = 12 y1 + n, x1 x2 + x3 x4 = y1 . If, instead of y1 , another root y2 or y3 of the resolvent cubic (17) had been employed in §48, quadratic equations different from (18) would have been obtained, such, however, that their four roots are x1 , x2 , x3 , x4 , paired in a new manner.