Set of real $a$ so that the inequality is defined but isn't true for a real $x$

Question

$$x(x-\sqrt {4+\log_a7})\lt \log_7 \frac a{49}$$

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.

Answer 1

$$x(x-\sqrt{4+\log_a7}) < \log_7 a - \log_7 49 =\frac1{\log_a 7} - 2$$

$$ x^2 - x\sqrt{4+\log_a 7} - \frac 1 {\log_a 7} + 2 < 0. $$

\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${}\,\ldots\,{}$]

Answer 2

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 $$2x^2-2\sqrt 6\ x+3=2\left(x-\frac{\sqrt 6}{2}\right)^2\ge 0.$$

Answer 3

Let's examine when an inequality $$ x(x-A)<B $$ has no solutions. It becomes $$ x^2-Ax-B<0 $$ and the discriminant should be nonpositive, so that the quadratic only takes on nonnegative values. This means $$ A^2+4B\le0 $$ Now $A=\sqrt{4+\log_a7}$ and $B=\log_7a-2$. By the change of basis formula, $\log_a7=1/\log_7a$, so we can set $k=\log_7a$. Note that a condition is $a>0$, $a\ne1$. Moreover we need $4+\log_a7\ge0$, but I'll return later to this.

Our condition now reads $$ 4+\frac{1}{k}+4k-8\le0 $$ or $$ \frac{4k^2-4k+1}{k}\le0 $$ that gives $k<0$, because the numerator is $(2k-1)^2$, which is nonnegative, or $k=1/2$.

Thus one condition is $\log_7a<0$, that is, $a<1$, or $a=\sqrt{7}$

Now we return to $$ 4+\frac{1}{k}\ge0 $$ If $k<0$, this means $4k+1\le0$, that is $k<-1/4$ or $0<a\le1/\sqrt[4]{7}$; it is satisfied for $k=1/2$. So the final answer is $$ 0<a\le\frac{1}{\sqrt[4]{7}}\text{ or } a=\sqrt{7} $$