Question

How do I solve: x^2 = kx + 1 for x by completing the square?

Answer 1

start with
\[x^2-kx=1\] then
\[(x-\frac{k}{2})^2=1+\frac{k^2}{4}=\frac{4+k^2}{4}\]

Answer 2

then
\[x+\frac{k}{2}=\pm\frac{\sqrt{4+k^2}}{2}\] and finally
\[x=\frac{-k\pm\sqrt{4+k^2}}{2}\]

Answer 3

oohh.. o.o thank you!