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Quartic Equation

Descartes Solution

A quartic equation has the form
x4 + bx3 + cx2 + dx + e = 0

I will illustrate the steps by which we can arrive at the solution of a biquadratic or quartic equation.
  1. Express x4 + bx3 + cx2 + dx + e = 0 as
    x4 + px2 + qx + r = 0.

  2. Let x4 + px2 + qx + r = 0 be equivalent to
    x4 + px2 + qx + r = (x2 + Ax + B)(x2 + Cx + D)

  3. You will get A = -C , B + D + AC = p, AD + BC = q and BD = r

  4. B + D - A2 = p, A(D - B) = q and BD = r
    From these frame a cubic equation in A.
    (B + D) = (p + A2)
    D - B = q/A
    DB = r
    (B+D)2 = (D-B)2 + 4DB
    (p + A2)2 = (q/A)2 + 4r
    (pA + A3)2 = q2 + 4rA2
    A6 + 2pA4 + p2A2 = q2 + 4rA2
    A6 + 2pA4 + (p2-4r)A2 - q2 = 0

  5. Solve the cubic equation A6 + 2pA4 + (p2-4r)A2 - q2 = 0
    You will get one real root.

  6. Find the value of D and B with the help of A.

  7. Substitute the value of A, C, B and D in the equations
    (x2 + Ax + B) = 0
    and (x2 + Cx + D)= 0 to get two quadratic equations.

  8. Solve the quadratic equations to get the roots.

Ferrari's solution

A quartic equation has the form
x4 + bx3 + cx2 + dx + e = 0

I will illustrate the steps by which we can arrive at the solution of a biquadratic or quartic equation.
  1. Add (px + q)2 to both sides of the equation
    x4 + bx3 + cx2 + dx + e = 0
    x4 + bx3 + (c+p2)x2 + (d+2pq)x + (e+q2) = (px + q)2
  2. Choose p and q such that both sides of the equation is a perfect square.
    (x2 + (b/2)x + k)2 = (px + q)2
  3. Equate the constants of
    x4 + bx3 + (c+p2)x2 + (d+2pq)x + (e+q2) = (px + q)2
    and
    (x2 + (b/2)x + k)2 = (px + q)2
    to get a cubic equation in k. This cubic equation is known as resolvent cubic.
  4. Find the values of p and q with the help of k1 = value of k.
  5. Get two equation from  x2 + (b/2)x + k1 = ±(px + q)
    x2 + (b/2 + p)x + k1+q = 0
    and x2 + (b/2 - p)x + k1-q = 0
  6. Solve the two equations to get the solution.


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