Set of real $a$ so that the inequality is defined but isn't true for a real $x$ $$x(x-\sqrt {4+\log_a7})\lt \log_7 \frac a{49}$$
I reach the interval $(0,1)$ after looking for the discriminant of the quadratic to be less than zero. However, the solution in the book is an interval that includes $\sqrt 7$, which by hand you can figure out is a solution. Am I wrong? Is it the book? Thanks.

$$x(x-\sqrt{4+\log_a7}) < \log_7 a - \log_7 49 =\frac1{\log_a 7} - 2$$
$$
x^2 - x\sqrt{4+\log_a 7} - \frac 1 {\log_a 7} + 2 < 0.
$$
\begin{align}
\text{discriminant} & = (4+\log_a 7) - 4\left( - \frac 1 {\log_a 7} + 2 \right) = \log_a 7 + \frac 4 {\log_a 7} - 4 \\[12pt]
& = \begin{cases}
\left( \sqrt{\log_a 7} - \dfrac 2 {\sqrt{\log_a 7}} \right)^2 & \text{if }\log_a 7>0, \\[10pt]
-\left( \sqrt{-\log_a 7} + \dfrac 2 {\sqrt{-\log_a 7}} \right)^2 & \text{if }\log_a 7<0.
\end{cases}
\end{align}
When the discriminant is negative then the inequality will have no solution, and that happens when $\log_a 7<0$, i.e. if $a<1$.
[To be continued, maybe${}\,\ldots\,{}$]
