Certainly! Let's work through the formation of equations whose roots are provided.
### Part a: Roots [tex]\(i\)[/tex] and [tex]\(-i\)[/tex]
To form the quadratic equation with roots [tex]\(i\)[/tex] and [tex]\(-i\)[/tex], we use the fact that the sum of the roots [tex]\((i + (-i) = 0)\)[/tex] and the product of the roots [tex]\((i \cdot (-i) = -1)\)[/tex]. The quadratic equation in its standard form [tex]\(x^2 - (sum\:of\:roots)x + (product\:of\:roots) = 0\)[/tex] becomes:
[tex]\[ (x - i)(x + i) = x^2 + 1 \][/tex]
Therefore, the equation is:
[tex]\[ x^2 + 1 = 0 \][/tex]
### Part b: Roots [tex]\(2 + i\)[/tex] and [tex]\(2 - i\)[/tex]
For this case, we will follow a similar process. The sum of the roots is [tex]\((2 + i) + (2 - i) = 4\)[/tex] and the product of the roots is [tex]\((2 + i)(2 - i) = 4 + 1 = 5\)[/tex]. Thus, we derive the quadratic equation:
[tex]\[ (x - (2 + i))(x - (2 - i)) = x^2 - 4x + 5 \][/tex]
So, the equation is:
[tex]\[ x^2 - 4x + 5 = 0 \][/tex]
### Part c: Roots [tex]\(1 - 3i\)[/tex] and [tex]\(1 + 3i\)[/tex]
Similarly, the sum of the roots is [tex]\((1 - 3i) + (1 + 3i) = 2\)[/tex] and the product is [tex]\((1 - 3i)(1 + 3i) = 1 + 9 = 10\)[/tex]. Thus, the quadratic equation is:
[tex]\[ (x - (1 - 3i))(x - (1 + 3i)) = x^2 - 2x + 10 \][/tex]
So, the equation is:
[tex]\[ x^2 - 2x + 10 = 0 \][/tex]
### Part d: Roots [tex]\(1 + i\)[/tex], [tex]\(1 - i\)[/tex], and [tex]\(2\)[/tex]
Finally, for three roots [tex]\(1 + i\)[/tex], [tex]\(1 - i\)[/tex], and [tex]\(2\)[/tex], the polynomial formed by the roots is given by:
[tex]\[ (x - (1 + i))(x - (1 - i))(x - 2) \][/tex]
First, let's find the quadratic factor involving the complex conjugates:
[tex]\[ (x - (1 + i))(x - (1 - i)) = (x - 1 - i)(x - 1 + i) \][/tex]
[tex]\[ = (x - 1)^2 - (i)^2 \][/tex]
[tex]\[ = (x - 1)^2 + 1 \][/tex]
[tex]\[ = x^2 - 2x + 1 + 1 \][/tex]
[tex]\[ = x^2 - 2x + 2 \][/tex]
Now, multiply this quadratic polynomial by [tex]\((x - 2)\)[/tex]:
[tex]\[ (x^2 - 2x + 2)(x - 2) \][/tex]
[tex]\[ = x^3 - 2x^2 + 2x - 2x^2 + 4x - 4 \][/tex]
[tex]\[ = x^3 - 4x^2 + 6x - 4 \][/tex]
Therefore, the equation is:
[tex]\[ x^3 - 4x^2 + 6x - 4 = 0 \][/tex]
In summary, the equations whose roots are given are:
1. For roots [tex]\(i\)[/tex] and [tex]\(-i\)[/tex]: [tex]\(x^2 + 1 = 0\)[/tex]
2. For roots [tex]\(2 + i\)[/tex] and [tex]\(2 - i\)[/tex]: [tex]\(x^2 - 4x + 5 = 0\)[/tex]
3. For roots [tex]\(1 - 3i\)[/tex] and [tex]\(1 + 3i\)[/tex]: [tex]\(x^2 - 2x + 10 = 0\)[/tex]
4. For roots [tex]\(1 + i\)[/tex], [tex]\(1 - i\)[/tex], and [tex]\(2\)[/tex]: [tex]\(x^3 - 4x^2 + 6x - 4 = 0\)[/tex]
