(x-5)3%3D0-degree.jpeg "(x2-4x+13)(x-5)3=0 Degree")
Solving the Cubic Equation: (x²-4x+13)(x-5)³ = 0
This equation presents a cubic equation, meaning it has a highest power of 3 for the variable x. To solve this, we can utilize the following steps:
1. Factorization
The equation is already partially factored. We can see that the equation is a product of two factors:
- (x² - 4x + 13)
- (x - 5)³
For the product to equal zero, at least one of the factors must be equal to zero.
2. Solving for x in each factor
a) (x² - 4x + 13) = 0
This quadratic equation does not factor easily. We can use the quadratic formula to find the solutions:
- x = (-b ± √(b² - 4ac)) / 2a
Where a = 1, b = -4, and c = 13.
- x = (4 ± √((-4)² - 4 * 1 * 13)) / 2 * 1
- x = (4 ± √(-36)) / 2
- x = (4 ± 6i) / 2 (where 'i' is the imaginary unit, √-1)
- x = 2 ± 3i
Therefore, the solutions for this factor are x = 2 + 3i and x = 2 - 3i. These are complex solutions.
b) (x - 5)³ = 0
We can solve this by taking the cube root of both sides:
- x - 5 = 0
- x = 5
This gives us a real solution, x = 5.
3. Conclusion
The solutions to the cubic equation (x²-4x+13)(x-5)³ = 0 are:
- x = 2 + 3i
- x = 2 - 3i
- x = 5
Therefore, the equation has one real solution (x = 5) and two complex solutions (x = 2 + 3i, x = 2 - 3i).