1 Answer | Add Yours
First, you'll have to provide an equation. You've provided an expression, for the moment. We'll transform the expression into an equation:
x^2-4xy+5y^2+2y-4 = 0
We'll create the following groups;
(x^2-4xy+4y^2) + (y^2+2y + 1)- 1 - 4 = 0
We notice that we've made some changes within the expression you've provided, namely:
5y^2 = 4y^2 + y^2
- we've added and subtracted the value of 1
We notice that creating the groups above, we've created, in fact, two perfect squares:
(x-2y)^2 + (y+1)^2 - 5 = 0
Now, we'll have to determine what are the integers which added to give 5?
1 + 4, 2 + 3, 3 + 2, 4 + 1
Let x - 2y = 1 and y + 1 = 2
x = 2y + 1
y = 1 => x = 3
The first integer pair which satisfies the given equation is (3,1).
Let x - 2y = sqrt2 and y + 1 = sqrt3
Since x and y are not integer, we'll move to the next possibility:
x - 2y = 2 and y + 1 = 1 => y = 0
x = 2
The next possible pair of integer numbers, that makes the equality to hold is (2;0).
Therefore, the pairs of integer numbers that satisfy the given equation are: (3,1) and (2;0).
We’ve answered 319,844 questions. We can answer yours, too.Ask a question