The equation kx^2+4kx+5=0, where a is a constant, has no real roots. Find the range of possible values of k.

First, if k=0 then the equation has no real roots since in this case we would have 5=0 (this is a proof by contradiction).
A quadratic equation of the form ax^2+bx+c=0 has no real roots if the discriminant b^2-4ac<0.
In this case, we have a=k, b=4k, and c=5. Inputting these values for a, b and c into this inequality we have
(4k)^2-4k5<0,expanding this we get16k^2-20k<0,and dividing by 4 gives us4k^2-5k<0.
We want to find the value of k, so we factorise the left hand side to give us k(4k-5)<0. We sketch the quadratic y=k(4k-5) to see where the curve is below the y axis, as this is the region where k(4k-5)<0.
We can see from the diagram that this equation is satisfied when 0<k<5/4.
Combining this with our earlier observation that k=0 gives no real solutions, we have 0<=k<5/4.