How to solve this quadratic inequality?

Question

My question: Find the values of q for which the quadratic equation $qx^2-4qx+5-q=0$ will have no real roots. Does this mean the discriminant has to be less than 0 in order to get no real roots? My workings: $b^2-4ac < 0$, then i substitute the $b$, $a$ and $c$ into the equation. So, $(-4q)^2-4(q)(5-q) < 0$ Which is $12q^2-20q < 0$ I then factorise it which is $4q(3q-5)<0$. But this is wrong, can someone please explain? The answer is $0<q<1$ Please help thank you!

Answer 1

Your proof is correct , you have made a small algebraic mistake.

Your $$12q^2−20q<0$$ should have been $$20 q^2-20 q <0$$

You will get the correct answer with this inequality.

Answer 2

You solved wrong

$$ (-4q)^2-4(q)(5-q) <0 \Rightarrow 20q^2-20q <0 $$

Answer 3

$$(-4q)^2-4q(5-q)=\color{red}{20}q^2-20q<0$$

Hence $$q^2-q<0$$

$$q(q-1) <0$$

$$0<q<1$$