The answer to the question "Find the range of values of ${a}$ satisfying $-3\lt \frac{x^2+ax-2}{x^2-x+1}\lt 2$" given in the book is
Since $x^{2}-x+1>0$ for all real $x$, the given set of inequalities may be expressed as $-3\left(x^{2}-x+1\right)<x^{2}+$ $a x-2<2\left(x^{2}-x+1\right)$
On simplification, we obtain \begin{array}{lll} 4 x^{2}+(a-3) x+1>0 & \text { and } & x^{2}-(a+2) x+4>0 \text { for which } \\ (a-3)^{2}-16<0 & \text { and } & (a+2)^{2}-16<0 \end{array} or $\quad-4<\mathrm{a}-3<4 \quad$ and $\quad-4<\mathrm{a}+2<4$
$\therefore \quad$ a lies between $-1$ and 7 as well as between $-6$ and $2$.
The range of values for a is $(-1,2)$.
I have a problem with the fifth line from the start of the block i.e. how can one get from $4 x^{2}+(a-3) x+1>0$ (say it equation $1$) to $(a-3)^{2}-16<0 $, assuming $x\in R?$
[Note that $(a-3)^{2}-16$ is discriminant of equation 1].
If I did the same for the quadratic $x^2-4x-77<0$ ($x\in R$) then I get $16+308<0$ which is absurd. Is the book wrong? If not why is right in that step? If wrong how will we solve the question?
Thank you, for reading my question.
The question as you quote it seems a little ill-posed to me. It should say something like: "Find all values $a$ such that $-3< \frac{x^{2}+ax-2}{x^{2}-x+1}<2$ for all real $x$."
If for a given $a$ the quadratic equation $4x^{2}+(a-3)x+1$ is strictly positive for all real $x$, then it cannot have any real roots. The only way a quadratic equation can have no real roots is precisely when the discriminant is negative.
In $x^{2}-4x-77$ there is no parameter $a$ to be chosen, so we cannot find any condition on $a$.