Question

find the zeros of the following function.
$f(x) = x^2 - 4x + 13$
$-2 pm 3i$
$2 - 3i$
$2 pm 3i$

Explanation:

Step1: Use quadratic formula

For \(ax^2 + bx + c = 0\), roots are \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b=-4\), \(c = 13\).

Step2: Calculate discriminant

\(b^2 - 4ac=(-4)^2-4\times1\times13 = 16 - 52=-36\).

Step3: Find roots

\(x=\frac{4\pm\sqrt{-36}}{2}=\frac{4\pm6i}{2}=2\pm3i\).

Answer:

\(2\pm3i\) (corresponding to the option "2 ± 3i")