Why is discriminant less than zero? the question is find the range of values of $c$ for which the expression $4x^2-4x+4c^2-8$ is non-negative for all real values of $x$.
I got the discriminant but I don't understand why it has to be less than zero. If it is non-negative, shouldn't it be positive and more than zero?
Can someone help me explain this, thanks :)
 A: Here it is easy to complete the square to obtain
$$(2x-1)^2+4c^2-9\ge 0$$
You ought to recognise your discriminant here. The minimum value of the left-hand side clearly occurs with $2x=1$ and you should be able to finish from here.

With $ax^2+bx+c$ you can complete the square cleanly by multiplying by $4a$ to obtain $$4a^2x^2+4abx+4ac=(2ax+b)^2+4ac-b^2$$
You just need to bear in mind that if $a$ is negative, multiplying by $4a$ changes the sign.
The form of the equation in this question meant that no fiddling was necessary. But the general case shows you where the discriminant comes in, and might help you to understand the sign.
A: Do not confuse the sign of the discriminant with the sign of the expression. A negative discriminant ensures that the expression has no real root, i.e. the expression keeps the same sign for all $x$.
The negative discriminant is not enough: the expression must be positive for at least one value of $x$, then it is positive for all. (It suffices that the leading coefficient be positive, which is the case here.)
$$\forall x:ax^2+bx+c>0\iff b^2-4ac<0\land a>0.$$
A: Another possible approach : consider the function $$f(x)=4x^2-4x+4c^2-8$$ $$f'(x)=8x-4$$ The derivative cancels for $x=\frac 12$ (which corresponds to a minimum) and, at this point, the value of the function is $$f(\frac 12)=4c^2-9$$ and you want this to always be non negative.
A: The Parabula is "smiling" and hence in order it will be positive, you need that there will be no intersection with the $x$ axis, that is there are no solutions to the equation
$$4x^2-4x+4c^2-8=0.$$
That happens if and only if the discriminant is negative.
