Solve u² + 49 = 0 and 2v² + 18 = 0. Give the positive number b for which the roots are ± bi in each…

Step 1. The first equation. $u^2 = -49$, so $u = \pm\sqrt{-49} = \pm 7i$.

Step 2. The second equation. $2v^2 = -18$, so $v^2 = -9$ and $v = \pm 3i$.

Step 3. Check. $(7i)^2 + 49 = -49 + 49 = 0$, and $2(3i)^2 + 18 = -18 + 18 = 0$.