To solve the equation ((x^2 + 2x + 2)(x^2 + 2x - 4) = -5), we first set:
Let (y = x^2 + 2x). Then, the equation can be rewritten as:
[
(y + 2)(y - 4) = -5
]
Now, expanding the left-hand side gives us:
[
y^2 - 4y + 2y - 8 = y^2 - 2y - 8
]
Setting the equation equal to (-5):
[
y^2 - 2y - 8 = -5
]
Now, add 5 to both sides:
[
y^2 - 2y - 3 = 0
]
Next, we can solve this quadratic equation using the quadratic formula:
[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
]
where (a = 1), (b = -2), and (c = -3):
[
y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)}
]
[
y = \frac{2 \pm \sqrt{4 + 12}}{2}
]
[
y = \frac{2 \pm \sqrt{16}}{2}
]
[
y = \frac{2 \pm 4}{2}
]
This results in two solutions for (y):
- (y = \frac{6}{2} = 3)
- (y = \frac{-2}{2} = -1)
Now, we will replace (y) back with (x^2 + 2x) and solve for (x):
- For (y = 3):
[
x^2 + 2x = 3
]
[
x^2 + 2x - 3 = 0
]
Factoring gives:
[
(x + 3)(x - 1) = 0
]
So, (x = -3) or (x = 1).
- For (y = -1):
[
x^2 + 2x = -1
]
[
x^2 + 2x + 1 = 0
]
This factors to:
[
(x + 1)^2 = 0
]
So, (x = -1) (with a multiplicity of 2).
Final Solutions
The solutions to the original equation ((x^2 + 2x + 2)(x^2 + 2x - 4) = -5) are:
[
x = -3, \quad x = 1, \quad x = -1 \text{ (with multiplicity 2)}
]
