<s>[INST] Set of values of $x$ 
If the inequality
  $$(1-a^2)x^2+(2a-3)x+1<0$$
  is true for all values of $a$ then the set of values of $x$ is?

I took two cases (the parabola opens upwards and the parabola opens downwards). For the first case the value of $x$ is all real except the interval containing roots. For the second case the value of $x$ is the interval containing the roots. I couldn't proceed with this as I couldn't eliminate $a$. Any ideas? Thanks
 [/INST] $$\left(1-a^2\right)x^2+(2 a-3) x+1<0$$
Expand and collect wrt $a$
$$-a^2 x^2+2 a x+\left(x^2-3 x+1\right)<0$$
Multiply both sides by $-1$ and change the $<$ into $>$
$$a^2x^2-2ax-\left(x^2-3 x+1\right)>0$$
This is true for any $a$ if and only if the discriminant $\Delta$ of the polynomial 
$P(a)=a^2x^2-2ax-\left(x^2-3 x+1\right)$ is negative
$\Delta=4x^2+4x^2(x^2-3x+1)=4x^2(x-2) (x-1)<0$
That is $\color{red}{1<x<2}$
Hope this is useful
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