(x%5E2%2B2x%2B4)%2B3-0.jpeg "(x^2+2x+3)(x^2+2x+4)+3 0")
Solving the Equation (x^2 + 2x + 3)(x^2 + 2x + 4) + 3 = 0
This problem involves simplifying and solving a quartic equation (an equation with the highest power of x being 4). Let's break it down step by step:
1. Substitution:
To simplify the equation, we can use a substitution. Let's define:
y = x² + 2x
Now, our equation becomes:
(y + 3)(y + 4) + 3 = 0
2. Expanding and Simplifying:
Expanding the product and combining terms, we get:
y² + 7y + 12 + 3 = 0 y² + 7y + 15 = 0
3. Solving the Quadratic Equation:
Now we have a quadratic equation in terms of 'y'. We can solve this using the quadratic formula:
y = (-b ± √(b² - 4ac)) / 2a
Where a = 1, b = 7, and c = 15
Plugging in these values:
y = (-7 ± √(7² - 4 * 1 * 15)) / 2 * 1 y = (-7 ± √(-23)) / 2 y = (-7 ± i√23) / 2
4. Back-Substituting:
Remember, y = x² + 2x. Let's substitute back to find the values of x:
- For y = (-7 + i√23) / 2:
x² + 2x = (-7 + i√23) / 2
This is a quadratic equation in terms of 'x'. We can solve it using the quadratic formula again, but the solutions will involve complex numbers.
- For y = (-7 - i√23) / 2:
x² + 2x = (-7 - i√23) / 2
Similarly, this will also lead to complex solutions for 'x'.
Therefore, the original equation (x² + 2x + 3)(x² + 2x + 4) + 3 = 0 has four complex solutions.
Important Note: This problem highlights the fact that not all equations have real-number solutions. In this case, we encounter complex solutions due to the negative value under the square root in the quadratic formula.