Show that equation x^2 + kx - 3 = 0 has distinct real roots for all real values of k

We know that for a quadratic equation ax^2 + bx + c = 0, discriminant D = b^2 - 4ac
The condition for 'distinct' real roots is D > 0. Thus, to prove that x^2 + kx - 3 = 0 has distinct real roots, we got to prove D > 0 for it.

Lets have a look at what D comes out for x^2 + kx - 3 = 0

D
= k^2 - 4(1)(-3)
= k^2 + 12

We obviously understand that if k is a real number (and its given in the question that it is), k^2 will always be greater than or equal to 0 (k^2 ≥ 0). The reason of it being the fact that square of any real number can never be a negative number and thus it is always greater than or equal to 0.

Now, if k^2 is always greater than or equal to 0, then, by common sense, we can say that k^2 + 12 must be greater than or equal to 12. We can understand the same in form of a mathematical inequality. We know that 

k^2 ≥ 0

By adding 12 to both sides of the inequality

=> k^2 + 12 ≥ 12

So, now that we know that k^2 + 12 is greater than or equal to 12, it automatically becomes greater than 0. Like it is obvious, isn't it? k^2 + 12 ≥ 12, which means that the least value it can take is 12. It is always going to be 12 or more than that, and thus we can say that k^2 + 12 is also > 0.

Now, we had found that D = k^2 + 12 for the equation in the question. we've found out that k^2 + 12 is always greater than 0. Hence, we can say that the equation will have distinct and real roots.