Find a quadratic function with real coefficients whose zeros include 1+the square root of 3i.
since the coefficients are all real, the complex roots must come in conjugate pair. So the minimum polynomial would be
y = (x-(1+√3 i))(x-(1-√3 i))
= ((x-1)-√3 i)((x-1)+√3 i)
= (x-1)^2 - (√3 i)^2
= x^2-2x+1 + 3
= x^2-2x+4
To find a quadratic function with real coefficients whose zeros include the complex number 1 + √3i, we need to use the complex conjugate.
The complex conjugate of 1 + √3i is 1 - √3i. This means that if 1 + √3i is a root, then 1 - √3i must also be a root.
Now, let's set up the quadratic function:
Let's call the quadratic function f(x), and the roots are 1 + √3i and 1 - √3i. Recall that for a quadratic function, if x is a root, then (x - a) is a factor of the quadratic function, where a is the root.
So, the factors of the quadratic function are (x - (1 + √3i)) and (x - (1 - √3i)). To find the quadratic function, we multiply these factors together:
f(x) = (x - (1 + √3i))(x - (1 - √3i))
Let's simplify this expression:
f(x) = (x - 1 - √3i)(x - 1 + √3i)
To simplify further, we can use the difference of squares formula, (a + b)(a - b) = a^2 - b^2:
f(x) = (x - 1)^2 - (√3i)^2
Notice that (√3i)^2 is -3, so we can simplify further:
f(x) = (x - 1)^2 - (-3)
Now, simplify the expression inside the parentheses:
f(x) = (x^2 - 2x + 1) + 3
Finally, combine like terms:
f(x) = x^2 - 2x + 1 + 3
f(x) = x^2 - 2x + 4
Therefore, the quadratic function with real coefficients whose zeros include 1 + √3i is f(x) = x^2 - 2x + 4.