Discriminant when coefficient of x^2 == 0.

Question

$f(x)=4kx^2+(4k+2)x+1$ So the discriminant is $b^2-4ac$ making

$(4k+2)^2-4(4k)(1)$

which simplifies to

$16k^2+16k+4-16k$

which simplifies to

$16k^2+4$

A question states that when k=0, f(x) cannot have two distinct roots, which I get as f(x) become $2x+1$ which is linear, and when you sub 0 into the discriminant it equals 4 which seems to suggest it has 2 real roots, why is this not the case?

$16(0)^2+4 = 4$

Answer 1

The concept of discriminant works when we are dealing with a quadratic equation and an equation is quadratic if and only if it is of the form $ax^2+bx+c$ with $a\ne 0$. So the moment you make $a=0$ you are not allowed to use the concept of discriminant.