use sqrt property to solve:
(4k-1)^2+28=1
k=??

Question

jim_thompson5910 · Answer

$$\large (4k-1)^2+28=1 $$



$$\large (4k-1)^2=1-28 $$



$$\large (4k-1)^2=-27 $$



$$\large 4k-1=\pm\sqrt{-27}$$



$$\large 4k-1=\sqrt{-27} \ \ \textrm{or} \ \ 4k-1=-\sqrt{-27}$$



$$\large 4k-1=3i\sqrt{3} \ \ \textrm{or} \ \ 4k-1=-3i\sqrt{3}$$



$$\large 4k=1+3i\sqrt{3} \ \ \textrm{or} \ \ 4k=1-3i\sqrt{3}$$



$$\large k=\frac{1+3i\sqrt{3}}{4} \ \ \textrm{or} \ \ k=\frac{1-3i\sqrt{3}}{4}$$

anonymous · Answer

oh hey!! the page didn't show this for me! You rock :)

jim_thompson5910 · Answer

yeah the messaging system is a bit weird for me too