9x^2 + 16y^2 - 36x + 96y + 36 = 0 how do i find the domain to this problem?
To find the domain of the given equation, we need to determine the set of possible values for the variables x and y that satisfy the equation.
Starting with the given equation: 9x^2 + 16y^2 - 36x + 96y + 36 = 0
We can rearrange the terms to separate the x and y variables:
9x^2 - 36x + 16y^2 + 96y + 36 = 0
Next, we can group the x terms together and complete the square to factorize the equation. For the x terms, we can factor out a 9:
9(x^2 - 4x) + 16y^2 + 96y + 36 = 0
Now, we need to complete the square inside the parentheses. To do this, we take half of the coefficient of x (-4/2 = -2) and square it (-2^2 = 4), then add and subtract this value inside the parentheses:
9(x^2 - 4x + 4 - 4) + 16y^2 + 96y + 36 = 0
Simplifying further:
9[(x - 2)^2 - 4] + 16y^2 + 96y + 36 = 0
Expanding the square:
9(x - 2)^2 - 36 + 16y^2 + 96y + 36 = 0
Combining like terms:
9(x - 2)^2 + 16y^2 + 96y = 0
Now we can factor out a common factor (9):
9[(x - 2)^2 + (16/9)y^2 + (96/9)y] = 0
Simplifying the fractions:
9[(x - 2)^2 + (16/9)(y^2 + 6y)] = 0
At this point, we can see that the equation is a sum of squares. In order for the whole equation to equal zero, each term must be zero.
(x - 2)^2 = 0
This tells us that x - 2 must equal zero, so x = 2.
Similarly, (y^2 + 6y) = 0. To solve this, we factor out a y:
y(y + 6) = 0
This gives us two possible solutions: y = 0 or y = -6.
Hence, the domain of the equation is x = 2 and y can be any real number.