Part (a)


Consider the discriminant of the quadratic polynomial x^2 - x + 1.


The discriminant is  d = b^2 - 4ac = (-1)^2 - 4*1*1 = 1 - 4 = -3.


We see that the discriminant is negative.

It means that the polynomial does not have real zeroes.  From the other side hand, 
its leading coefficient at x^2, "1", is positive. 
It means that the polynomial  x^2 - x + 1 is always positive, for all real values of x.


Thus the statement (a) is proved and part (a) is completed.



                Part (b)


Consider this rational function  .


As we proved in part (a), its denominator is always positive, for all real values of x.


Therefore, while solving inequality 

     < 2,     (1)

we can multiply both its sides by the positive factor x^2 - x + 1, without flipping the inequality sign.


By doing it, we get the equivalent inequality

    x^2 + ax - 2 < 2*(x^2 - x + 1).     (2)


Simplify it step by step

    x^2 + ax - 2 < 2x^2 - 2x + 2,

    0 < x^2 - (2-a)x + 4,

    x^2 - (2-a)x + 4 > 0.               (3)


Thus, inequality (1) is held for all real x if and only if inequality (3) is held for all real x.


To find, what it implies for "a", let's consider the discriminant of the quadratic polynomial in the left side of (3).
It is

    D = b^2 - 4ac = (-(2-a))^2 - 4*4 = (2-a)^2 - 16.


Inequality (3) is held for all real x if and only if discriminant D is positive

    D > 0,  or  (2-a)^2 - 16 >0,  or  (2-a)^2 > 16,  or  |2-a| > ,  or  |2-a) > 4.    (4)


The set of solutions to inequality (4) are all values of "a" that are remoted from 2 farther than 4 units.

So, the set of "a" when inequality (1) is held for all real x, is the union  (,) U (,).


ANSWER.  The set of values "a", such that inequality   < 2  is satisfied 

         for all real values of x is the union  (,) U (,).