Rational Root Test

Date: 10/28/96 at 21:44:54
From: Tybon Wu
Subject: Solving Quadratic-Quadratic Systems Algebraically

The question is:
Solve the following system:

Equation 1: x^2 + y^2 - 6y - 1 = 0
Equation 2: xy + 6 = 0

I tried to solve Equation 2 for y (y = -6/x) and substitute into 
Equation 1.  Now I am having trouble solving the x variable.  Help!

Date: 01/04/97 at 18:54:57
From: Doctor Yiu
Subject: Re: Solving Quadratic-Quadratic Systems Algebraically

Dear Tybon,

You have made a good start by putting y = -6/x into Equation 1.
However, you should note that by dividing Equation 2 by x, you are 
assuming that x is not equal to 0 since you can't divide by 0, so 
you have to remember to consider the case where x = 0 separately.  
If x = 0, equation #2 becomes 6 = 0, which is impossible, so there 
are no solutions with x = 0.  So, you are fine in assuming x is not 
equal to zero, and we can continue with the problem.

Substitution of y = -6/x into Equation 1 gives:

x^2 + (36/x^2) + (36/x) - 1 = 0

Clearing denominators:

   x^4 + 36 + 36x - x^2 = 0
   x^4 - x^2 + 36x + 36 = 0 

---------------------------------

Solving higher degree equations generally involves some guesswork.  If 
you can factor the polynomial, the question will be much easier.  In 
the present case, note that (x+1) divides the sum of the first two 
terms, namely, x^4 - x^2, and also the sum of the last two terms, 
namely, 36x + 36.  (x+1) is a factor of the polynomial so we can write 
the following:

   x^2(x^2-1) + 36(x+1) = 0
x^2(x+1)(x-1) + 36(x+1) = 0
   (x+1)[x^2(x-1) + 36] = 0
  (x+1)(x^3 - x^2 + 36) = 0

---------------------------------

Here, we need another factorization, that of the cubic x^3 - x^2 + 36.
By the RATIONAL ROOT TEST, you know that you only need to try 
fractions with: 

(1) Numerators that evenly divide the constant term 36
(2) Denominators that evenly divide the leading coefficient 1

In other words, here, we need only try: x = 1, 2, 3, 4, 6, 9, 12, 18, 
36, and the corresponding NEGATIVE numbers.

Beginning with the smaller values, after some trial and error, we find
x = -3 a root:  

(-3)^3 - (-3)^2 + 36 = -27 - 9 + 36 = 0 

This means that x+3 is a factor, and, by division: 

x^3 - x^2 + 36 = (x+3)(x^2  -  4x  + 12)

---------------------------------

Back to the 4th degree equation in x, we have:

(x+1)(x+3)(x^2-4x+12) = 0

This gives x = -1, -3. The roots of the quadratic:

x^2 - 4x + 12 = 0

are imaginary, because the discriminant (-4)^2 - (4)(1)(12) = -32
is negative.

---------------------------------

This means that there are only two REAL solutions:

With x = -1, we have y = 6
With x = -3, we have y = 2

---------------------------------

If you also need the complex solutions, then from the quadratic 
equation:

x^2 - 4x + 12 = 0

you get, either by the quadratic formula or by completing squares:

x = 2 + or - Sqrt{-8) = 2 + or -2 Sqrt(-2)

If x = 2 + 2Sqrt(-2), then: 
    
   y = -6/x
      =  -3/(1+Sqrt(-2)) 
      = -3(1-Sqrt(-2))/((1+Sqrt(-2))(1-Sqrt(-2)) 
      = -3(1-Sqrt(-2))/3 
      =  -(1 - Sqrt(-2))
      =  -1 + Sqrt(-2)

Similarly, from x = 2 - 2Sqrt (-2), the value of y is  -1 - 2Sqrt(-2)

---------------------------------

The complete solution consists of four points, each with an
x-coordinate and a y-coordinate which are the points of intersection, 
the solutions, of the system consisting of Equations 1 and 2: 

      (-1, 6)
      (-3, 2)
      (2 + 2Sqrt(-2), -1 + Sqrt(-2))
      (2 - 2Sqrt(-2), -1 - Sqrt(-2)) 

-Doctor Yiu,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/