Finding the value of q where parabolas don't intersect?

Question

Find the value of q where parabolas $f(x)=2x^2+qx+1$ and $g(x)=-x^2+2x-2$ don't intersect.

How should I approach this? All I know is that the discriminant is supposed to be less than zero.

Answer 1

Let a point of intersection be $(a,b)$. Then $$f(a)=g(a) \implies 2a^2+aq+1=-a^2+2a-2 \implies 3a^2+(q-2)a+3=0.$$ Can you continue from here?

Answer 2

$-4<q<8$

indeed to intersect must be

$2x^2+q x+1=-x^2+2 x-2$

$3 x^2+(q-2) x+3=0$

which must have no solutions and this happens when the discriminant is negative

$(q-2)^2-36<0\to q^2-4q-32<0\to \color{red}{-4<q<8}$