E-StoreHOW TO SOLVE A CUBIC EQUATION :
THEORY OF EQUATION
SOLUTION OF A CUBIC EQUATION_____
- dipendra kumar misra
INTRODUCTION :
Well what do we do when we get a cubic equation. Either we try to see if x=1,-1 is the solution and then factorise or we leave it saying it cannot be solved.
But many of you will not be knowing that any cubic equation can be solved.
The first solution was presented by Gerolamo Cardano in his 1545 book Ars Magna.(Ref : wikipedia).
Now lets begin with the main part.
CUBIC EQUATION
CARDANO’s SOLUTION
A general cubic equation has the form :
x3 + ax2 + bx + c =0 ……….1)
Now as a wise mathematician we would first reduce the equation before solving it. So what we do is introduce a new variable :
x = t + h …..2)
So that we can collect the coefficient of some term in t and put it equal to 0 which will yield the value of h.
(t+h)3 + a(t+h)2 + b(t+h) + c = 0
t3 + h3 + 3th(t+h) + a(t2 + h2 + 2th) + b(t+h) + c=0
t3 + t2(3h + a) + t(3h2 + 2ah + b) + (h3 + c + bh + ah2) = 0
now see the best way is to disappear t2 since we can easily get the solution of h that is
h = -a/3……………3)
in all other we will get a quadriatic or cubic both of which will increase the complexity so we disappear the term t2 .
So we get :
t3 + t(b – 2/3a2 + a2/3) + (c – ab/3 + a3/9 – a3/27) = 0
t3 + t(b – a2/3) + (c + 2a3 /27 – ab/3) = 0……4)
Now we define
p = b – a2/3
q = c + 2a3/27 – ab/3
so we get
t3 + pt + q = 0…………5)
Enjoy the style in which Cardano simplifies the equation before solving it. In this way the mathematics is born.
Now we will need to substitute a new way which works only for cubic. It is to put
t = u + v………6)
Now it is not clear why we use it but it will soon be clear. Inserting it we get :
(u + v)3 + p(u+v) + q =0
u3 + v3 + 3uv(u+v) + p(u+v) + q = 0
u3 + v3 + (3uv + p)(u+v) + q = 0
Now we all know what to substitute : not u=-v but :
uv = -p/3…….7)
v = -p/3u
u3 + (-p/3u)3 + q = 0
u3 - p3/(27u3) + q = 0
u6 – p3/27 + qu3 = 0……….8)
Now it is a very simple problem of quadratic equation.
Put u3 = L and we get
L2 + qL – p3/27 = 0………9)
Now I am not presenting the solution which is devilishly lengthy but I hope the solution is clear.
WHAT TO DO NOW?
After solving the above equation we get the value of L and hence of u and then using the result 7) we get the value of v.
So we get the value of t which is
t = u + v
hence we get our 3 roots which can be found by using the result 2 and 3 to get
x = u + v – a/3
the value of p and q can be converted to our original a,b,c to get the solution.Cheers!!!
: Blade-X
dated// 8th july,2009
PS : Nugorama I think this one was not complicated
PSS : Hey aren’t the days getting bored on earth.
For more detail see the article under wikipedia : cubic equation.
comment which should be the next article :
1 : Euler-Cauchy EquiDimensional Differential eqn
2 : Newton-Raphson Root Finding Method
Comments (3)
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