How to solve
x^2-y^2-2y-1
We "solve" equations, yours is not an equation.
Did you perhaps want to factor it ??
x^2 - (y^2 + 2y + 1)
= x^2 - (y+1)^2
ahhh, a difference of squares ...
= (x+y+1)(y-y-1)
Oh okay thank you
To solve the equation x^2 - y^2 - 2y - 1 = 0, we can follow these steps:
Step 1: Write the equation in a more standard form.
Rearrange the equation by grouping the y terms together:
x^2 - (y^2 + 2y) - 1 = 0
Step 2: Complete the square for the y terms.
We want to complete the square for the expression inside the parentheses, y^2 + 2y. To do that, we need to add and subtract the square of half of the coefficient of y, which is (2/2)^2 = 1, inside the parentheses:
x^2 - (y^2 + 2y + 1 - 1) - 1 = 0
This can be further simplified to:
x^2 - [(y + 1)^2 - 1] - 1 = 0
Step 3: Simplify the equation.
By expanding the expression and simplifying, we obtain:
x^2 - (y + 1)^2 + 1 - 1 - 1 = 0
x^2 - (y + 1)^2 - 1 = 0
Step 4: Rearrange the equation to isolate the square term.
Move the constant term (-1) to the other side of the equation:
x^2 - (y + 1)^2 = 1
Step 5: Take the square root of both sides.
To solve for y, we need to isolate the square term by itself, so we take the square root of both sides:
√(x^2 - (y + 1)^2) = √1
The right side simplifies to ±1.
Step 6: Solve for y.
By taking the square root, we consider both the positive and negative square root solutions:
y + 1 = ±1
Solve for y by subtracting 1 from both sides of the equation:
y = -1 ± 1
This gives us two possible values for y:
1) y = -1 + 1 = 0
2) y = -1 - 1 = -2
Therefore, the solutions to the equation x^2 - y^2 - 2y - 1 = 0 are y = 0 and y = -2.