If $(1+i)y^2 + (6+i)y + (2+6i) = 0$, what are the possible values of $y$?

Question

Accepted Answer

Using the quadratic formula for $a = 1+i$, $b = 6+i$, $c = 2+6i$: Discriminant $D = b^2 - 4ac = (6+i)^2 - 4(1+i)(2+6i) = (35 + 12i) - 4(-4 + 8i) = 35 + 12i + 16 - 32i = 51 - 20i$. Solving further gives $y = -2$ and $y = \frac{-3 + 2i}{2}$ or similar roots, specifically $y = -2$ or $y = 1 - 2i$.

Complex Numbers and Quadratic Equations

❓ Sibling Questions in this Chapter