<s>[INST] Find all values of the parameter $a$ so that $f(x)= y$ is always positive Find all values of the parameter $a$ so that $y>0$. 
The equation is:
$$y = (a − 1)x^2 − (a + 1)x + a + 1$$
Thanks!
 [/INST] First of all, the discriminant should be negative but $a-1>0$.
We get
$$(a+1)^2 - 4(a-1)(a+1) < 0 \Longleftrightarrow a^2+2a+1 - (4a^2 - 4) = -3a^2+2a +5 < 0.$$
By using quadratic formula
$$-3a^2+2a+5 = (-3a +5 )(a + 1) < 0.$$
The formula above will have negative value when $a < -1$ or $a > \frac{5}{3}$.
Because the value of $a$ should be greater than 1, so that
$$a > \frac{5}{3}.$$
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