Cubic Equation Question...?

Question

For a standardized cubic equation, x^3+px+q=0, how many roots does it have? (is there a delta thing around like quadratic equation to help resolve the problem?)

I know a formula, it's x=A^1/3+B^1/3， but doesn't it only lead to ONE root? how to get the other two?(if there are others), and how do I know if the roots are real or imaginary? THX

Answer 1

For a standard cubic equation there will be three (3) roots. There is no standard formula like quadratic equation, you have to use the nullification technique i.e. keep putting an arbitary number to see the equation becomes 0. once it is zero then work to find out others

an example will do the trick

let's assume an equation

x^3+x^2+x-3=0

(now we see by putting a value of x=1 satisfies the equation. Thus x-1 is a root of this equation)

Now we get
x^3+x^2+x-3=0
or, x^3-x^2+2x^2-2x+3x-3=0
or, x^2(x-1)+2x(x-1)+3(x-1)=0
or, (x-1)(x^2+2x+3)=0

now usning quadratic equation formula for (x^2+2x=3) we get

x = [-2 +/- sqrt (4-12)]/2
or, x = -1 +/- 1/2 (sqrt-8)
or, x = -1 +/- sqrt (-2)
or, x = -1 +/- √2 i [i stands for sqrt(-1), which is imaginary]

Thus the assumed equation has three roots namely
1, -1-√2 i, -1+√2 i (the last two are imaginary roots)
-------------------------------

if x = A^1/3+B^1/3

then,
x^3 = A+B+3A^1/3B^2/3+3A^2/3B^1/3
(cubing both sides)

now
x^3-(A+B+3A^1/3B^2/3+3A^2/3B^1/3)=0
is the equation where exponents of x^2 and x are zero and mathematically it has three roots. The other two roots to be found the same way I did before ... like this

M(x-A^1/3-B^1/3)+N(x-A^1/3-B^1/3)+O(x-A^1/3-B^1/3)=0

now you will get this

(M+N+O)(x-A^1/3-B^1/3)=0

now solve M+N+O=0 using common quadratic formula equation and BINGO! you get the other two roots ...

cool :)

Answer 2

it quite is fairly complicated in comparision with the quatratic as quickly as.... there's a protracted-length formulation for this... Now, the thought in the back of this formulation is using some linear transformation to the equation to finally end up with the style ok*X^3 + L*X + N = 0 (may be performed with all cubic equations). Then, for the concepts one proceeds utilising the roots of team spirit. especially, we factorize x^3 - 3*a*b + a^3 + b^3, i.e. x^3 - 3*a*b + a^3 + b^3 = (x + a + b)*(x + a*o +b*o^2)*(x + b*o +a*o^2) the place o = -a million/2 + i*sqrt(3)/2 i.e a complicated third root of team spirit Now, the complicated roots of x^3 - 3*a*b + a^3 + b^3= 0 are patently: x = - (a + b), x = - (a*o +b*o^2) and x = - (b*o + a*o^2) concepts (I) Then, we learn: x^3 - 3*a*b + a^3 + b^3= 0 with ok*X^3 + L*X + N = 0 or equivalently X^3 + (L/ok)*X + (N/ok) = 0. From this we receive: L/ok = - 3*a*b and N/ok = a^3 + b^3. We proceed via fixing this simulatneous equation to locate a, b in terms of ok, L, M and replace to concepts (I) to get some new concepts. to locate the suitable concepts of our preliminary equation we word the inverse preliminary linear tranformation to the concepts.... and voila the final cubic equation is solved... As you comprehend fixing your equation demands many lof of algebraic calculations . . . it quite is why the cubic formulation is a lot lots larger than the quatratic...

Answer 3

Yes as the person atop me stated there are 3 roots and the way to find them depends.

Say examples it is x^3-x^2=0 then simply factor x out
x(x^2-1)=x(x-1)(x+1)

If it is something like Ax^3+Bx^2+Cx+D where A,B,C,D<0 and A,B,C,D>0, then you have to use the guess and check factor method.Where all the coefficients of f(x) are put in a straight line, and and all multiples of +-(A)(D) must be checked so that the multiple say it is called f, such that ((f(f(Af)-B))-C))-D=0

The method I suggested above is quite laboursome but if your in High School, you'll probably get quite easy problems but sometimes in the professional mathematics world degree >2 polynomials become very hard to find. Thanfully there is a very close and approximative method, not a whole value but very close to the actual solution, it is called the Newton root finding method, it is quite hard to master and understand and probably not at all required for High School students.

Answer 4

If you can find one root, you can reduce the cubic to a 2nd degree by long division, then find the roots by quadratic equation if they are not apparent. They will be complex if b^2 - 4ac < 0.

Wikipedia has an excellent description of the method for finding all the roots of a cubic, but it can be quite tedious.

Answer 5

For this question, I will use an cubic equation question as an example. Anyway the general formula is px^3+qx^2+rx +s = 0
where p,q,r and s are constant.

Example: 2x^3+3x^2-11x-6 = 0.
We let 2x^3+3x^2-11x-6 = (x-@)(ax^2+bx+c)
@*c = 6 (constant term)

Therefore possible values of @ are +1,-1,+2,-2,+3,-3,+6 and -6.

Let f(x) = 2x^3+3x^2-11x-6 and test the possible values untill we obtain f(@) = 0.

Thus, f(1) = 2+3-11-6 is not equals to 0.
f(-1) = -2+3+11-6 is not equals to 0.
f(2) = 16+12-22-6 = 0. and therefore @=2

and so, 2x^3+3x^2-11x-6 = (x-2)(ax^2+bx+c)
= (x-2)(2x^2+bx+3)....(coeff.of x^3,x^0)
= (x-2)(2x^2+7x+3)...(coeff.of x^2)
= (x-2)(2x+1)(x+3)....(coeff.of x^2)

so when f(x)=0, x=2,x= -1/2 , x= -3 .

Answer 6

There is a formula to get the solution for any cubic equation:
(see http://www.math.vanderbilt.edu/~schectex/courses/cubic/)

If you want to find it for the depressed cubic (like the one you have given), you can use your formula to find the first root (let's call it a).
Then you divide the polynomial by (x-a) and you get a quadratic equation that you can easily solve.

A cubic equation always has at least 1 real root.
The other 2 roots can be real or complex.
(Complex roots always come in pairs (c+di and c-di).

Answer 7

first of all a general cubic equation must be of the form..
x^3+p x^2+qx+c=0.
IT will have 3 roots.

Answer 8

you can use u^3-v^3=q and uv=p/3,...........or you can use the sum of the roots,starting with (x-x1)(x-x2)(x-x3)=x^3+px+q=0 (x1,x2,x3 are the roots of the equation),leading to x1+x2+x3=0, x1x2+x2x3+x3x1=p and x1x2x3=-q.