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.

For $a=\sqrt 7$ one has
$$x\left(x-\sqrt{4+\log_{\sqrt 7}7}\right)\lt \log_7{\frac{\sqrt 7}{49}},$$
i.e.
$$x(x-\sqrt 6)\lt -\frac 32,$$
i.e.
$$2x^2-2\sqrt 6\ x+3\lt 0.$$
However, this is not true because
$$2x^2-2\sqrt 6\ x+3=2\left(x-\frac{\sqrt 6}{2}\right)^2\ge 0.$$
