Answer:
[tex]k=\pm3[/tex]
Step-by-step explanation:
[...] if you can write the LHS as a perfect square, or if you can't spot a factorization of it right away, if and only if the discriminant [tex]\Delta = b^2-4ac[/tex] (or, if b is an even number, 1/4 of it) is zero.
I see it! I see it!
Stare at it for a while. First term is [tex]3(2x)^2[/tex], third term is [tex]3(1)^2[/tex], we are missing a double product, but we can play with k. For the LHS to be [tex]3(2x\pm1)^2 = 3(4x^2\pm4x+1) = 12x^2\pm12x+3[/tex] you just need [tex]4k= \pm12 \rightarrow k=\pm3[/tex].
I don't see it...
Then number crunching it is. Set the discriminant to 0, solve for k
[tex]\frac{\Delta}4 = 4k^2-12\cdot 3 =0 \rightarrow 4k^2=36 \\k^2 = 9 \rightarrow k=\pm3[/tex]
