Quote:
Originally posted by Down With The King:
Find the other roots of this equation
x^3+2x^2+6x+3 |
Let this expression be equal to ax³ + bx² + cx + d. (Thus a = 1, b = 2, c = 6, and d = 3.)
First we must find the "depressed cubic" within the cubic by applying the substitution x = y - b / 3a.
This arrives at
ay³ + (c - b² / 3a)y + (d + 2b³/27a² - bc / 3a) =
y³ + (6 - 4/3)y + (3 + 16/27 - 4) =
y³ + 14y/3 - 11/27 = 0.
This gives an equation of the form y³ + Ay = B (A = 14/3, B = 11/27).
Now we must find numbers s and t such that 3st = A and s³ - t³ = B. Solving the first equation for s, s = A / 3t, we see that (A / 3t)³ - t³ = B, so t^6 + Bt³ - A³/27 = 0. If we substitue u = t³, this becomes u² + Bu - A³/27 = 0, which is quadratic and has solution
u = -B ± sqrt(B² + 4A³/27) / 2 =
(-11/27 ± sqrt(121/729 + 10976/729)) / 2 =
(-11/27 ± sqrt(121 + 10976)/27) / 2 =
(-11 ± sqrt(121 + 10976)) / 54 =
(-11 ± sqrt(11097)) / 54 =
(-11 ± 9 sqrt(137)) / 54.
We'll take the positive root, so u = (-11 + 9 sqrt(137)) / 54 = 1.74707....
Thus, as u = t³, t = cubrt(u) = cubrt((-11 + 9 sqrt(137)) / 54) = 1.20440....
We then know that s = A/(3t), so s = 14/9t = 1.29156....
It has been shown that s - t is a root of the "depressed cubic", so s - t = 0.087159... is a root of y³ + 14y/3 - 11/27 = 0.
Thus, x = y - b / (3a) = 0.087159... - 2/3 = -0.57950... is a root of the equation.
Dividing this solution via synthetic division from the initial expression, the other two roots can be found from the equation x^2 + (1.42049...)x + (5.17681...) = 0.
Using the quadratic formula, we can solve for x with x =
((-1.42049...) ± sqrt((-1.42049...)² - 4(5.17681...))) / 2 =
((-1.42049...) ± sqrt(-18.68945...)) / 2 =
-0.71024... ± i * sqrt(18.68945...) / 2 =
(-0.71024...) ± i(2.16157...).
Thus, it is solved. The roots are (-0.57950...) and (-0.71024...) ± (2.16157...)i.
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[ February 19, 2005, 04:12 PM: Message edited by: The Minish Link ]