Question

if f(x)=(x^2-8x+9)(x^2-6x+13) has four complex solutions written in factored form as f(x)=(a+bi)(a-bi)(c+di)(c-di). What is the sum of a, b, c, and d?

Answer 1

no, it should be
\[ f(x)=(x-(a+bi))(x-(a-bi))(x-(c+di))(x-(c-di))\]

Answer 2

you can solve both of these equations easily by completing the square
first one is
\[x^2-8x+9=0\]
\[x^2-8x=-9\]
\[(x-4)^2=-9+16=7\]
\[x-4=\pm\sqrt{7}\]
\[x=4\pm\sqrt{7}\] both roots are real not complex

Answer 3

this one
\[x^2-6x+13=0\] has complex roots, they are
\[x^2-6x=-13\]
\[(x-3)^2=-13+9=-4\]
\[x-3=\pm 2i\]
\[x=3\pm2i\]

Answer 4

you can still add up the roots,
you get \(3+3+4+4=14\)

Answer 5

Thank You :)