ax2 + bx + c = 0
which are then solved, either by factoring and solving for x, or - very often - by using the quadratic formula:
x = [-b + {b2 - 4ac}1/2]/ 2a
But what about solving basic cubic equations? Here we solve equations of the form:
x3 – px2 + qx – r = 0
Example:
x3 – x2 - x + 2 = 0
What I’ll show here, or rather demonstrate, is an algorithm for simple solution provided the numerical coefficient of the cubic term is 1.
We begin by writing: p = 1, q = -1, r = -2
And let:
D = [(- 4p3 r – 27r2 + 18 pqr – 4 q3 + p2 q2)] 1/2
With:
x1 = [p3 –9/2 (pq – 3r) – 3/2 x (-3)1/2 x D ]1/3
x2 = (p2 – 3q)/ x1
The ROOTS are then – in turn:
a1 = 1/3(p + r2 x x1 + r x x2)
WHERE:
r = - 1/2 + (-3)1/2/2 x (-3)1/2 D
r2 = -1/2 – (-3)1/2/2
Last TWO roots:
a2 = 1/3( p + r2 x x1+ r x x2)
a3 = 1/3( p + r x x1 + rx r2 x x2)
For the original eqn. in question, one obtains:
D = 7.681i
x1 = 0.578 + 1.001i
x2 = 1.73 – 2.997i
and
a1 = 1.103 - 0.665i
a2 = 1.103 + 0.665i
a3 = -1.206
IF the Roots are correct THEN:
a1 + a2 + a3 = 1
and:
(a1 x a2) + (a 2 x a3) + (a3 x a1) = q
AND
a1 x a2 x a3 = r
AND:
(a1 – a2) x (a2 – a3)x (a3 – a1) = D
We will just check the first: a1 + a2 + a3 = 1
We have then:
[1.103 - 0.665i] + [1.103 + 0.665i] + (-1.206) = 2.206 - 1.206 = 1
Energetic readers can check the others!
Energetic readers can check the others!
No comments:
Post a Comment