Let's substitute u = x + 2 to simplify the equation:
With this substitution, our equation becomes:u^4 + u^2 - 12 = 0
This is a quadratic equation in u. Let's solve for u using the quadratic formula:
u = [-b ± √(b^2 - 4ac)] / 2a
In this case, a=1, b=1, and c=-12.
Plugging those values into the formula:
u = [-1 ± √(1^2 - 4(1)(-12))] / 2(1)u = [-1 ± √(1 + 48)] / 2u = [-1 ± √49] / 2u = [-1 ± 7] / 2
Therefore, we have two possible solutions for u:u = (7 - 1) / 2 = 6 / 2 = 3u = (-1 - 7) / 2 = -8 / 2 = -4
Now, substitute back x + 2 for u:For u = 3, x + 2 = 3x = 1
For u = -4, x + 2 = -4x = -6
Therefore, the solutions to the equation (x+2)^4 + (x+2)^2 - 12 = 0 are x = 1 and x = -6.
Let's substitute u = x + 2 to simplify the equation:
With this substitution, our equation becomes:
u^4 + u^2 - 12 = 0
This is a quadratic equation in u. Let's solve for u using the quadratic formula:
u = [-b ± √(b^2 - 4ac)] / 2a
In this case, a=1, b=1, and c=-12.
Plugging those values into the formula:
u = [-1 ± √(1^2 - 4(1)(-12))] / 2(1)
u = [-1 ± √(1 + 48)] / 2
u = [-1 ± √49] / 2
u = [-1 ± 7] / 2
Therefore, we have two possible solutions for u:
u = (7 - 1) / 2 = 6 / 2 = 3
u = (-1 - 7) / 2 = -8 / 2 = -4
Now, substitute back x + 2 for u:
For u = 3, x + 2 = 3
x = 1
For u = -4, x + 2 = -4
x = -6
Therefore, the solutions to the equation (x+2)^4 + (x+2)^2 - 12 = 0 are x = 1 and x = -6.