Question

Find a number k such that the given equation has exactly one real solution. x^2-kx+4=0

Answer 1

k=4

since
x^2-4k+4=(x-2)^2=0
which gives solution x=2 multiplicity 2

Answer 2

or you might like this better...

Answer 3

to complete the square we do
\[(-k/2)^2\]
but we want this to be 4
so set them equal and solve for k

Answer 4

\[(\frac{-k}{2})^2=4\]

Answer 5

\[\frac{-k}{2}= \pm 2 => -k =\pm 4 => k= \mp 4\]

Answer 6

so -4 would also have worked