Question

The equation 12x² + 4kx + 3 = 0 has real and equal roots, if

Answer 1

Answer:

$k=\pm3$

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 $\Delta = b^2-4ac$ (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 $3(2x)^2$, third term is $3(1)^2$, we are missing a double product, but we can play with k. For the LHS to be $3(2x\pm1)^2 = 3(4x^2\pm4x+1) = 12x^2\pm12x+3$ you just need $4k= \pm12 \rightarrow k=\pm3$.

I don't see it...

Then number crunching it is. Set the discriminant to 0, solve for k

$\frac{\Delta}4 = 4k^2-12\cdot 3 =0 \rightarrow 4k^2=36 \\k^2 = 9 \rightarrow k=\pm3$