Rational Root TestDate: 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/ |
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