How many ordered pairs of integers (x,y) are there such that 2x2−2xy+y2=225?
using the quadratic equation, I solved for y
where a = 1 , b = -2x and c = 2x^2-225
y = (2x ± √(4x^2 - 4(4x^2 - 225) )/2
= (2x ± √(900 - 20x^2)/2
= x ± √(225 - x^2)
now we want x and y to be integers, so
225 - x^2 must be a perfect square
so possible results for 225- a perfect square giving us a perfect square are
225 - 0 = 225 , good one
225 - 1
225 - 4
225 - 9
225 - 16
225 - 25
225 - 36
225 - 49
225 - 64
225 - 81 = 144 ahhhh
225 - 100
225 -121
225 - 144 = 81 --- another one
225 - 169
225 - 196
225- 225 = 0 --- that one works
so we have x = 0, ±9, ±12, ±15
each one will give an integer for y
(0, ±15) , (9, 21), (9, -3), (-9, 3), (-9, -21) ....
so I count 14 such ordered pairs
brilliant says its wrong... 14 is wrong... it meant 2x^2−2xy+y^2=225
To find the number of ordered pairs of integers (x, y) that satisfy the equation 2x^2−2xy+y^2=225, we can use a process called completing the square.
First, let's rearrange the equation as follows:
2x^2 - 2xy + y^2 = 225
(x^2 - 2xy + y^2) + x^2 = 225
(x - y)^2 + x^2 = 225
Now, we can see that this equation has the form of a Pythagorean triple, where (x - y) and x are the two legs of a right triangle, and 225 is the square of the hypotenuse.
Let's express 225 as a square number:
225 = 15^2
Substituting this back into the equation, we have:
(x - y)^2 + x^2 = 15^2
Since we are looking for ordered pairs of integers, we can create a table of perfect squares up to 15^2:
1^2 = 1
2^2 = 4
3^2 = 9
...
13^2 = 169
14^2 = 196
15^2 = 225
We can see that there are 15 perfect squares less than or equal to 15^2.
Now, we can set up two equations and solve for (x - y) and x:
(x - y)^2 = 1 => x - y = ±1
x^2 = 225 => x = ±15
For every combination of the values of (x - y) and x, we have a unique ordered pair (x, y). Thus, there are 2 options for (x - y) and 2 options for x, giving a total of 2 * 2 = 4 ordered pairs.
Therefore, there are 4 ordered pairs of integers (x, y) that satisfy the equation 2x^2−2xy+y^2=225.