Finding the values of $q$ for which the quadratic equation $qx^2-4qx+5-q=0$ will have no real roots. Find the values of $q$ for which the quadratic equation $qx^2-4qx+5-q=0$ will have no real roots. 
So I've gotten as far as using the discriminant to find the values of $q$, but I'm stuck on the last step.
$$(-4q)^2-4(q)(5-q)<0$$
$$16q^2-4q(5-q)<0$$
$$16q^2-20q+4q^2<0$$
$$20q^2-20q<0$$
$$20q(q-1)<0$$
What step should be taken next to find the values of $q$?
Thank you.
 A: You need to solve the quadratic inequality in $q$. First start by sketching the curve $f(q) = q(q-1)$, with $q$ being the horizontal axis and $f(q)$ the vertical axis. (We drop the coefficient of $20$ as it doesn't affect the shape of the curve). There will be two $q$-intercepts, namely $0$ and $1$. The parabola's ends will be upward pointing with the portion of the curve between the two intercepts lying below the axis. Hence the region you require is between the two $q$-intercepts, i.e. $0 < q < 1$.
A: This is where it's handy to make a sign chart. 
First, let's consider when it is zero:
$$20q(q-1) = 0 \Leftrightarrow q = 0 \textrm{ or } q = 1$$
Now, let's see what happens on the intervals $(-\infty,0),(0,1),(1,\infty)$. I claim that we can plug in any point within those intervals and whatever sign it has, that will be the sign on the whole interval. Let's accept that for now and see what it tells us.
For $(-\infty,0)$: take $q = -1$ then $20q(q-1) = -20(-2) = 40 > 0$ so this does not work 
For $(0,1)$: take $q = 1/2$ then $20q(q-1) = 10(-1/2) < 0$ so this works. 
For $(1,\infty)$, take $q = 2$ then $20q(q-1) > 0$ so this does not work. 
Thus, the interval we seek is $(0,1)$. 
The reason this works depends on some calculus. Without going into details, we are dealing with a continuous function. Intuitively, that means when drawing $20q(q-1)$ on a graph, you can do so without lifting your pencil. So imagine a sign change: it would have to cross the $x$-axis at some point, i.e. have a zero. Thus for a continuous function, identifying its roots is identifying precisely where the sign changes. 
A: You can also take this approach.  We cannot have $ \ q \ = \ 0 \ $ since this would "collapse" the expression to $ \ 5 \ = \ 0 \ $ , which ceases to be an equation. (The requirement of no real roots is "trivially" met because there are no solutions at all.  Martin Sleziak commented on this above while I was still getting this edited. )  This means we can divide the equation through by $ \ q \  $ to produce
$$ x^2 \ - \ 4x \ + \ (\frac{5}{q} - 1 ) \ = \ 0 \ \ . $$
Upon completing the square, we obtain
$$ (x^2 \ - \ 4x \ + \ 4) \ + \ (\frac{5}{q} - 1 - 4 ) \ = \ 0 \ \ \Rightarrow \ \ (x \ - \ 2)^2 \ = \ 5 \ - \ \frac{5}{q} \ \ , $$
which we can see has no real-valued solutions for $ \ 5 \ - \ \frac{5}{q} \ < \ 0 \ \ \Rightarrow \ \ 1 \ < \ \frac{1}{q} \ \ \Rightarrow \ \  1 \ >  \ q \ $ .  This establishes the solution interval $ \ 0 \ < \ q \ < \ 1 \ $ .  The right-hand side of this equation is positive for $ \ q \ < \ 0 \ $ , so the original quadratic equation does have real roots for this case.  Hence, $ \ 0 \ < \ q \ < \ 1 \ $ is the only solution interval.
We can also see this as locating the vertex of a parabola $ \ y \ = \ qx^2 \ - \ 4qx \ + \ (5 - q) \ $ so that the curve has no $ \ x-$ intercepts.  For $ \ q \ > \ 0 \ $ , the parabola is $ \ y \ = \ x^2 \ - \ 4x \ + \ (\frac{5}{q} - 1 ) \ $ or $ \ y \ - \ (\frac{5}{q} - 5 ) \ = \ (x \ - \ 2)^2 \   $ , which locates its vertex at $ \ ( \ 2 \ , \  \frac{5}{q} - 5 \  ) $ .  This parabola "opens upward", so it has no $ \ x -$ intercepts for $ \ \frac{5}{q} - 5 \ > \ 0 \ $ , which again gives the solution interval $ \ 0 \ < \ q \ < \ 1 \ $ .  For $ \ q \ < \ 0 \ $ , which corresponds to a "downward-opening" parabola,
$$ \ y \ = \ qx^2 \ - \ 4qx \ + \ (5 + |q|) \ $$  
$$ \Rightarrow \ \  y \ = \ -x^2 \ + \ 4x \ + \ \frac{5}{|q|} \  + \ 1  \ \ \text{or} \ \ \ y \ - \ (\frac{5}{|q|} + 5 ) \ = \ -(x \ - \ 2)^2 \ \ ,  $$   
the vertex is found to lie at $ \ ( \ 2 \ , \  5 \ + \ \frac{5}{|q|}  \  ) $ , which means that this parabola unavoidably has $ \ x-$ intercepts.  
