I don’t understand what my professor is asking. my original question and answer: 15t 2 + 2t+5 = 3t(t 2 +1). Let’s call this is equation (1) We’ll…

please help me with this. I don’t understand what my professor is asking….

my original question and answer:

15t2 + 2t+5 = 3t(t2+1). Let’s call this is equation (1)

We’ll have:               (1) 15t2 + 2t + 5 = 3t3 + 3t

15t2 + 2t + 5 – 3t3 – 3t = 0

-3t3 + 15t2 – t + 5 = 0

(t-5) (-3t2-1) = 0

t -5 = 0 or -3t2-1 =0

t=5 or 3t2+1 = 0 (divide by -1)

Consider 3t2 + 1. We have t2 always greater or equal than 0, so 3t2 must be greater or equal than 0. Therefore, 3t2+1 must be greater than 0. Therefore, there is no value of t for the equation 3t2 + 1.

Overall, there is only one value of t =5 that corrects to the (1) equation.

Her reply:

You are correct that there is only one real number solution of the equation, and that is t = 5. However, the exercise asks for all complex solutions, not just real number solutions.

So please complete the problem by also solving the equation 3t2 + 1 = 0 for the complex number solutions. Thanks!