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