Example Problem

Solve the following system:

x4 - 14x2 + y2 + 2x + 35 = 0
x - y - 1 = 0 		(1)
(2)

Solution

At first glance, it does not seem we are able to solve this system with our existing knowledge of algebra. However, let´s try to simplify the system first. Then, we will see what else we can do to solve it.

Using equation (2) gives us

y = x - 1 (3)

Substituting (3) into (1), we have:

x4 - 14x2 + (x - 1)2 + 2x + 35 = 0
x4 - 13x2 + 36 = 0 		(4)

Any equation in the form ax2n + bxn + c = 0 is said to be in quadratic form and can be solved using the quadratic formula or any other appropriate method. In this case, it appears that the factoring method can easily solve the equation:

x4 - 13x2 + 36 = (x2 - 9)(x2 - 4) = 0

Therefore, we have:

x2 - 9 = 0
x2 - 4 = 0 		(5)
(6)

From (5), we have:

x1 = 3
x2 = -3 		(7)
(8)

From (6), we have:

x3 = 2
x4 = -2 		(9)
(10)

Substituting (7) into (3), we obtain:

y1 = x - 1 = 2 (11)

Substituting (8) into (3), we obtain:

y2 = x - 1 = -4 (12)

Substituting (9) into (3), we obtain:

y3 = x - 1 = 1 (13)

Substituting (10) into (3), we obtain:

y4 = x - 1 = -3 (14)

Therefore, the solutions are:

x1 = 3
y1 = 2 		x2 = -3
y2 = -4 		x3 = 2
y3 = 1 		x4 = -2
y4 = -3