Question

roblem 3 Find the solutions to the general cubic a x^3 +b x^2+c x +d=0 and the solutions to the general quartic a x^4+b x^3+c x^2+d x+e=0. Remember to put a space between your letters. The solutions to the general quartic goes on for two pages it is a good idea to maximize your page to see it. It is a theorem in modern abstract algebra that there is no solution to the general quintic in terms of radicals.

Please write it clearly! thank you!

Answer 1

Solution for Cubic Equation :

Let's say you have the equation ax³ + bx² + cx + d = 0. The first thing to do is to get rid of the "a" out in front by dividing the whole equation by it. Then we get something in the form of x³ + ex² + fx + g = 0. The next thing we do is to get rid of the x² term by replacing x with (y - e/3). The new equation y³ + py + q = 0 has no y² term. This is much easier to solve, although it's still kind of hard and it took a lot of really smart people a long time to do.

Let's introduce two new variables, t and u, defined by the equations

             u - t = q               and   
             tu = (p/3)3.

If you're really interested, you might want to know what t and u actually are in terms of p and q:

         -3qSqrt{3} +- Sqrt{4p3 + 27 q2}
     t = --------------------------------- 
                     6 Sqrt{3}
                
           
          3qSqrt{3} +- Sqrt{4p3 + 27q2}
     u =  --------------------------------
                     6 Sqrt{3}

That +- is either + or - for BOTH of the equations at the same time. You can't have one as a + and the other as a - .

Anyway, y₁ = CubeRoot{t} - CubeRoot{u} will be a solution of y³ + py + q = 0.

Verify this result now (DON'T use the explicit formulas for t and u, just plug CubeRoot{t} - CubeRoot{u} into the cubic polynomial), and make sure you see why it works. To find the other two solutions of our cubic polynomial (if there are any real ones) we could divide y³ + py + q by its known factor (y - CubeRoot{t} + CubeRoot{u}), getting a quadratic equation that we could solve by the quadratic formula. Finally, once we have the solution values of y, we can find x = y - e/3.

As you can see, the complexity is much greater than in the Quadratic Formula. To try to go backward and come up with a closed form for the Cubic Formula in terms of the original a, b, c, d would be a real pain.

Solution for Quartic Equations :

Let the equation be : a' x⁴ + b' x^{​​​​3} + c' x² + d' x + e' = 0

Now divide the above equation with a' on both sides,

Assume coefficient of x³ (b'/a') as 'a'

Similarly assume coefficients of x² and x as 'b' and 'c' respectively

Let constant (e'/d') be 'd'.

The new simplified equation will be :

x⁴ + a x³ + b x² + c x + d = 0.

There are several ways to solve quartic equations. They all start by dividing the equation by the leading coefficient, to make it of the form

x⁴ + a x³ + b x² + c x + d = 0.

There is an easy special case if d = 0, in which case you can factor the quartic into x times a cubic. The roots then are 0 and the roots of that cubic. (To solve the cubic equation, see above.)

Substitute x = y - a/4, expand, and simplify, to get

y⁴ + e y² + f y + g = 0

where

e = b - 3 a²/8,
f = c + a³/8 - a b/2,
g = d - 3 a⁴/256 + a² b/16 - a c/4.

At this point, there are two useful special cases.

1. If g = 0, then, again, you can factor the quartic into y times a cubic. The roots of the original equation are then x = -a/4 and the roots of that cubic with a/4 subtracted from each.
2. If f = 0, then the quartic in y is actually a quadratic equation in the variable y². Solve this using your favorite method, and then take the two square roots of each of the solutions for y² to find the four values of y which work. Subtract a/4 from each to get the four roots x.

Henceforth we can assume that d, f, and g are all nonzero.

First Solution Method

Here is one way to proceed from this point. The related auxiliary cubic equation

z³ + (e/2) z² + ((e²-4 g)/16) z - f²/64 = 0

has three roots. Find those roots by solving this cubic (again, see above). None of these is zero because f isn't zero. Let p and q be the square roots of two of those roots (any choices of roots and signs will work), and set

r = -f/(8 p q).

Then p², q², and r² are the three roots of the above cubic. More important is the fact that the four roots of the original quartic are

x = p + q + r - a/4,
x = p - q - r - a/4,
x = -p + q - r - a/4, and
x = -p - q + r - a/4.

This solution was discovered by Leonhard Euler (1707-1783).

Second Solution Method

Here is another way to proceed from the above point. The quartic in y must factor into two quadratics with real coefficients, since any complex roots must occur in conjugate pairs. Write

y⁴ + e y² + f y + g = (y² + h y + j) (y² - h y + g/j)

Since f and g aren't zero, neither j nor h is zero either. Without loss of generality, we can assume h > 0 (otherwise swap the two factors). Then, equating coefficients of y² and y,

e = g/j + j - h²,
f = h (g/j - j).

so

g/j + j = e + h²,
g/j - j = f/h.

Adding and subtracting these two equations,

2 g/j = e + h² + f/h,
2 j = e + h² - f/h.

Multiplying these together,

4 g = e² + 2 e h² + h⁴ - f²/h².

Rearranging,

h⁶ + 2 e h⁴ + (e²-4 g) h² - f² = 0.

This is a cubic equation in h², with known coefficients, since we know e, f, and g. We can use the solution for cubic equations shown above to find a positive real value of h², whose existence is guaranteed by Descartes' Rule of Signs, and then take its positive square root. This value of h will give a value of j from 2 j = e + h² - f/h. Once you know h and j, you know the quadratic factors of the quartic in y. Notice that you do not need all the roots h² of the cubic, and that any positive real one will do. Further notice that h² = 4 z from the previous section.

Once you have factored the quartic into two quadratics, finishing the finding of the roots is simple, using the Quadratic Formula. Once the roots y are found, the corresponding x's are gotten from x = y - a/4.

Let's take one example and solve it for better understanding :

Example :

Find the roots of x⁴ + 6 x³ - 5 x² - 10 x - 3 = 0.

Solution :

• Set x = y - 3/2, substitute and expand,

  y⁴ + (-37/2) y² + 32 y + (-231/16) = 0,
  e = -37/2, f = 32, g = -231/16,
  h⁶ + (-37) h⁴ + 400 h² - 1024 = 0.

  Solving this cubic in h², we find

  h² = 16, h² = (21 + sqrt[185])/2, h² = (21 - sqrt[185])/2.

  We only need one of these, so we pick h² = 16, so

  h = 4,
  j = (e + h² - f/h)/2,
  j = -21/4,

  and one factorization is

  (y² + 4 y - 21/4) (y² - 4 y + 11/4) = 0,

  or

  (x² + 7 x + 3) (x² - x - 1) = 0.

  From here the rest is easy, and

  x = (-7 +- sqrt[37])/2, x = (1 +- sqrt[5])/2,

  are the four roots of the original equation.