Hello, JaysFan31!

Let f be the complex polynomial: f(x) .= .(3 + i)x² + (-2 - 6i)x + 12

Calculate the following values: f(i) and f(2 + 3i)

I calculated these values as such:

f(i) .= .(3 + i)i^2 + (-2 - 6i)i + 12

. . Doing the math: .f(i) .= .15 - 3i

f(2 + 3i) .= .(3 + i)(2 + 3i)(2 + 3i) + (-2 - 6i)(2 + 3i) + 12

. . Doing the math: .f(2 + 3i) .= .-1 + 13i

Is this right? . . . . They look good to me!

Also, it then asks find a complex number z

such that the equation f(x) = z has a unique solution.

(Use the Quadratic Formula. .How do you do this?

We have: .(3 + i)x² + (-2 - 6i)x + 12 .= .z

Or: . (3 + i)x² + (-2 -6i)x + 12 = z .= .0

We have a quadratic equation with: a = (3 + i), b = (-2 - 6i), c = (12 - z)

. . . . . . . . . . . . . . . . . . . . . . . . . _____________________

. . . . . . . . . . . . . . . . -(-2 - 6i) ± √(-2 - 6i)² - 4(3 + i)(12 - z)

Then we have: . x .= . ----------------------------------------------

. . . . . . . . . . . . . . . . . . . . . . . . . . 2(3 + i)

A quadratic will have one solution if its discriminant equals zero.

That is: . (-2 - 6i)² - 4(3 + i)(12 - z) .= .0

Now solve for z . . .