Have a quad day

For a polynomial (in the form $ax^2+bx+c$ ) to have equal zeroes, the discriminant must be equal to zero.

$\therefore b^2-4ac=0 \\ (-6)^2-4(3)(k)=0 \\ 36-12k=0 \\ k=\frac{36}{12}=\boxed{3}$

Sum of roots = $\dfrac {-(-6)}{3} = 2$

Since the roots are equal, The roots are $\text {1 and 1}$

Product of roots = $\dfrac {k}{3} = 1 \times 1 =1$

Thus, $k =3$

divide by 3 and we have $x^2 - 2x + \left( \dfrac{k}{3} \right)$ That begins with the first two terms of the well known square of bynomial $(x-1)^2 = x^2 - 2x + 1$ . So just set the third term equal to $1$ and you're done. So $k = 3$ .