2004-07-30, 00:35 | #1 |
Sep 2002
2·331 Posts |
Equality of quadratic equations
Does anybody know what math techniques are available to solve
two quadratic equations for equality ? The only one I am aware of is substituting in ordered pairs until the result of the two equations are equal. An example: What ordered pair y,x solves y^2 + 1393y + 463 = x^2 + 6x + 407 What (if any) direct methods exist ? There is also the issue of how many solutions 0, 1, 2+, infinite ? If the answer involves software, what program is required ? If it involves a branch of mathematics, what one(s) ? I tried graphing (in a spreadsheet), but didn't find any solutions, maybe there is a specific technique to use that I'm not aware of. Different software ? By plugging in the following ordered pair a solution is found, y = 3550 x = 4186 a solution with both quadratics equal = 17548119 There is at least one more solution for this particular example. |
2004-07-30, 02:44 | #2 |
Jul 2004
16_{8} Posts |
Check out Dario Alpern's "Generic Two Integer Variable Equation Solver":
http://www.alpertron.com.ar/QUAD.HTM He offers a lot of background information about this subject. BTW: You might not have typed your equation correctly because your solution doesn't appear to work. See the following solution generated by his applet. ============= APPLET OUTPUT (STEP-BY-STEP MODE) ==================== x^2 - y^2 + 6 x - 1393 y - 56 = 0 by Dario Alejandro Alpern First of all we must determine the gcd of all coefficients but the constant term, that is: gcd(1, 0, -1, 6, -1393) = 1. Dividing the equation by the greatest common divisor we obtain: x^2 - y^2 + 6 x - 1393 y - 56 = 0 We try now to solve this equation module 9, 16 and 25. There are solutions, so we must continue. We want to convert this equation to one of the form: x´^2 + B y^2 + C y + D = 0 Multiplying the equation by 4: 4 x^2 - 4 y^2 + 24 x - 5572 y - 224 = 0 4 x^2 + ( 24)x + ( - 4 y^2 - 5572 y - 224) = 0 To complete the square we should add and subtract: ( 6)^2 Then the equation converts to: ( 2 x + 6)^2 + ( - 4 y^2 - 5572 y - 224) - ( 36) = 0 ( 2 x + 6)^2 + ( - 4 y^2 - 5572 y - 260) = 0 Now we perform the substitution: x´ = 2 x + 6 This gives: x´^2 - 4 y^2 - 5572 y - 260 = 0 Multiplying the equation by -1: - x´^2 + 4 y^2 + 5572 y + 260 = 0 -x´^2 +( 4 y^2 + 5572 y) + 260 = 0 -x´^2 +((-2)^2 y^2 + 2*(-2)*(-1393) y) + 260 = 0 Adding and subtracting (-1393)^2: -x´^2 +((-2)^2 y^2 + 2*(-2)*(-1393) y + (-1393)^2) + 260 - (-1393)^2 = 0 -x´^2 +( - 2 y - 1393)^2 - 1940189 = 0 Making the substitution y´ = - 2 y - 1393: - x´^2 + y´^2 - 1940189 = 0 ( y´ + x´) ( y´ - x´) = 1940189 Now we have to find all factors of 1940189. Since 1940189 is equal to 1 times 1940189, we can set: y´ + x´ = 1 y´ - x´ = 1940189 x´ = -970094 y´ = 970095 x = -485050 y = -485744 and also: x = 485044 y = -485744 and also: x = -485050 y = 484351 and also: x = 485044 y = 484351 Since 1940189 is equal to 163 times 11903, we can set: y´ + x´ = 163 y´ - x´ = 11903 x´ = -5870 y´ = 6033 and also: x = -2938 y = -3713 and also: x = 2932 y = -3713 and also: x = -2938 y = 2320 and also: x = 2932 y = 2320 |
2004-07-30, 09:03 | #3 |
Sep 2002
2×331 Posts |
Thank You
I did have a typing error, + 463 should have been + 469 |
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