Is it possible for a cubic function to have only 2 real distinct roots, each of order 1? Explain why or why not. Include examples.
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No, a cubic equation will have triple roots. Why? in an equation of x^n +...=constant, there will be n roots.
Yes, it is possible for a cubic function to have only 2 real distinct roots, each of order 1. To understand why, let's first define what we mean by "order" of a root.
The order of a root refers to how many times that root appears in the polynomial equation. For example, if a root has an order of 1, it means it appears only once in the equation. If a root has an order of 2, it means it appears twice, and so on.
To determine the number and order of roots of a cubic function, we can consider its behavior and the fundamental theorem of algebra. The fundamental theorem of algebra states that a polynomial of degree n will have exactly n roots, including both real and complex roots.
Now, for a cubic function, which is a polynomial of degree 3, it can have a maximum of 3 roots, including both real and complex roots. However, it is possible for a cubic function to have fewer than 3 distinct roots.
In the case where a cubic function has only 2 real distinct roots, it means that the other root is a complex root. Since complex roots always come in conjugate pairs, the complex root will have an order of 2.
Here's an example to illustrate this:
Consider the cubic function f(x) = x^3 - 2x^2 + 2x - 1. This function has only 2 real distinct roots: x = 1 and x = -1. These roots each have an order of 1 because they appear only once in the equation. The complex root is x = i, a complex number, and its conjugate is x = -i. These complex roots also have an order of 1, as they only appear once in the equation.
In summary, a cubic function can have only 2 real distinct roots, each of order 1, when the third root is a complex root of order 2. The fundamental theorem of algebra guarantees that a cubic function will have exactly 3 roots, so if only 2 real roots are present, the missing root must be a complex root.