To solve the equation ((y^2 + 2y + 4)^2 - 7(y^2 + 2y + 4) + 12 = 0), let's first introduce a substitution to simplify the expression. Let:
[
x = y^2 + 2y + 4
]
Now, we can rewrite the equation in terms of (x):
[
x^2 - 7x + 12 = 0
]
Next, we will factor the quadratic equation. We want to find two numbers that multiply to (12) and add to (-7). These numbers are (-3) and (-4). Therefore, we can factor the equation as:
[
(x - 3)(x - 4) = 0
]
Setting each factor to zero gives:
[
x - 3 = 0 \quad \Rightarrow \quad x = 3
]
[
x - 4 = 0 \quad \Rightarrow \quad x = 4
]
Now, we substitute back (x = y^2 + 2y + 4).
Case 1: (y^2 + 2y + 4 = 3)
[
y^2 + 2y + 4 - 3 = 0 \quad \Rightarrow \quad y^2 + 2y + 1 = 0
]
[
(y + 1)^2 = 0 \quad \Rightarrow \quad y + 1 = 0 \quad \Rightarrow \quad y = -1
]
Case 2: (y^2 + 2y + 4 = 4)
[
y^2 + 2y + 4 - 4 = 0 \quad \Rightarrow \quad y^2 + 2y = 0
]
[
y(y + 2) = 0
]
[
y = 0 \quad \text{or} \quad y + 2 = 0 \quad \Rightarrow \quad y = -2
]
Final Solutions
So, the solutions to the original equation are:
[
y = -1, \quad y = 0, \quad y = -2
]
These are the final results!
