Answer
$\left\{-4,4\right\}$
Work Step by Step
$x^2-kx+4=0$ is the standard form $ax^2+bx+c=0$
We have $a=1,b=-k$ and $c=4$.
For repeated real solution the discriminant must be zero.
$ b^2-4ac=0$
Substitute the values of $a, b, $ and $c$.
$(-k)^2-4(1)(4)=0$
Simplify.
$k^2-16=0$
Add $16$ to both sides.
$k^2-16+16=0+16$
Simplify.
$k^2=16$
Use square root property.
$k=\pm \sqrt{16}$
Use $\sqrt{16}=4$.
$k=\pm 4$
Hence, the values of $k$ are $-4$ and $4$.
