Sure, let's solve the equation step by step:
First, we start with the given equation:
[tex]\[ x^2 - 2x - 2y - y^2 = 0 \][/tex]
To solve for [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex], we recognize that this is a quadratic equation in terms of [tex]\(y\)[/tex]. Let's rewrite it in the standard quadratic form:
[tex]\[ y^2 + 2y + (2x - x^2) = 0 \][/tex]
This can be viewed as a standard quadratic equation [tex]\(a y^2 + by + c = 0\)[/tex] where:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 2\)[/tex]
- [tex]\(c = 2x - x^2\)[/tex]
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex] and [tex]\(c\)[/tex] into the formula:
[tex]\[ y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (2x - x^2)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-2 \pm \sqrt{4 - 4(2x - x^2)}}{2} \][/tex]
[tex]\[ y = \frac{-2 \pm \sqrt{4 - 8x + 4x^2}}{2} \][/tex]
[tex]\[ y = \frac{-2 \pm \sqrt{4x^2 - 8x + 4}}{2} \][/tex]
Notice that the expression under the square root can be factored:
[tex]\[ 4x^2 - 8x + 4 = (2x - 2)^2 \][/tex]
So, the equation becomes:
[tex]\[ y = \frac{-2 \pm \sqrt{(2x - 2)^2}}{2} \][/tex]
Since the square root of a square is the absolute value, we have:
[tex]\[ \sqrt{(2x - 2)^2} = |2x - 2| \][/tex]
Thus, our solutions can be written as:
[tex]\[ y = \frac{-2 \pm |2x - 2|}{2} \][/tex]
Now consider the two cases for the absolute value:
1. When [tex]\(2x - 2 \ge 0 \)[/tex], i.e., [tex]\(x \ge 1\)[/tex]:
[tex]\[ y = \frac{-2 + (2x - 2)}{2} = \frac{2x - 4}{2} = x - 2 \][/tex]
2. When [tex]\(2x - 2 < 0\)[/tex], i.e., [tex]\(x < 1\)[/tex]:
[tex]\[ y = \frac{-2 - (2x - 2)}{2} = \frac{-2 - 2x + 2}{2} = \frac{-2x}{2} = -x \][/tex]
Consequently, we have two solutions:
[tex]\[ y = -x \quad \text{or} \quad y = x - 2 \][/tex]
So, the solution to the equation is:
[tex]\[ y = -x \quad \text{or} \quad y = x - 2 \][/tex]
