Equality of quadratic equations

[dsouza123]

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.

[jfollas]

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

[dsouza123]

Thank You

I did have a typing error, + 463 should have been + 469