Ask anyone who took freshman
math, and they’ll tell you of their introduction to algebra - which is one of
the most useful branches of mathematics, since it paves the way for properly
doing derivatives, and integrations in calculus. And almost everyone taking it
would have been exposed to quadratic equations of the form:
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
(and let interested readers check the others):
We have then:
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
(and let interested readers check the others):
We have then:
[1.103 - 0.665i] + [1.103 + 0.665i] + (-1.206) = 2.206 -
1.206 = 1
All of the above will be found to check out for the given equation!
All of the above will be found to check out for the given equation!
PROBLEMS:
1. For the example problem above, check that the roots
obtained satisfy each of the following:
a) (a1 x a2) +
(a
2 x a3)
+ (a3
x a1)
= q
AND
b) a1 x a2 x a3 = r
AND
b) a1 x a2 x a3 = r
2. Using the procedure formulated in the article, solve the
cubic equation:
x3 – 6x2 + 11x – 6
Again, check to ensure the roots you obtained are the
correct ones!
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