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## 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.