Download Real Roots of Cubic Equation

Real Roots of Cubic Equation
(Tomas Co, 10/7/2014)
Objective:
Given a cubic equation, + + + = 0, find the condition when the
equation would yield three real roots, and find the roots.
Method:
Use the Tschirnhaus-Vieta approach.
Let
3 − 2 − 9 + 27 =
and =
27
3
1. Condition for all roots being real: < 0
(if > 0, only one real roots, see page 4 for this case)
2. Evaluation of the roots:
= 2−
3
3
= arccos !=−
3
Then the roots are
1
"# = cos + !
3
1
" = cos % + 2&' + !
3
1
" = cos % + 4&' + !
3
Example:
Consider the equation
− 10 + 7 + 1 = 0
Then = −26.3333, = −49.7407, = 5.9255, = 0.2966 and ! = 3.3333
Yielding the roots: {9.2299, −0.1215,0.8917}.
1
Derivation of Formulas:
1. First, reduce the cubic equation to a “depressed cubic” equation.
-
-
Let = 0 + !, then the original equation becomes
+ 2!
! + ! + 0 + 3! + 0 + 3! +
0 + 1! +
2=0
By setting ! = −/(3), the coefficient of 0 becomes zero, yielding
0 + 0 + = 0
where = (3 − )/(3 )and = (2 − 9 + 27 )/(27 ).
(1)
2. Next, consider the following trigonometric identity
Or
cos(36) = cos(26) cos(6) − sin(26) sin(6)
= (cos(6) − sin(6) ) cos(6) − (2 sin(6) cos(6)) sin(6)
= cos(6) − 3 cos(6) sin(6)
= cos(6) − 3 cos(6) (1 − cos(6) )
cos(36) = 4 cos(6) − 3 cos(6)
4 cos(6) − 3 cos(6) − cos(36) = 0
3. Transform the reduced cubic (1) to match the trigonometric identity (2).
Let 0 = cos(6) and substitute to (1) to obtain
cos(6) + cos(6) + = 0
Then multiply the equation by 4/ to match the first term in (2),
4 cos(6) +
4
4
cos(6) + = 0
To continue to match with (2), we need
4
=
−3
→
=
2−
3
This yields,
4 cos(6) − 3 cos(6) −
2
3
=0
(2)
So comparing with identity (2), we conclude that
cos(36) =
3
Due to the periodicity of cosines, we have three distinct solutions for 6 if we let
= arccos 3
and set 36# = , 36 = + 2& and 36 = + 4&, or
cos(6# ) = cos + 2&
+ 4&
;cos(6 ) = cos ;cos(6 ) = cos 3
3
3
4. Transform back to the desired solution for .
That is, the roots are
= 0 + ! = cos(6) + !
+!
3
+ 2&
" = cos(6 ) + ! = cos +!
3
+ 4&
" = cos(6 ) + ! = cos +!
3
"# = cos(6# ) + ! = cos Note: Since = 2:−/3, we need < 0 to have be real. Further, with real we
will also have = arccos(3/()) be real.
3
If ; > <, only one Real Root:
̅ = 2
3
3
> = asinh 1 2
̅
>
"# = −2 sinh 1 2
3
3
4