Find the values of $b$ for which the equation $2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$ has only one solution

Solution 1:

You have $$x^2+(4+b)x+16=0\tag1$$ This is correct.

However, note that when we solve $$2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$$ we have to have $$bx+28\gt 0\quad\text{and}\quad 12-4x-x^2\gt 0,$$ i.e. $$bx\gt -28\quad\text{and}\quad -6\lt x\lt 2\tag2$$

Now, from $(1)$, we have to have $(4+b)^2-4\cdot 16\geqslant 0\iff b\leqslant -12\quad\text{or}\quad b\geqslant 4$.

Case 1 : $b\lt -14$

$$(2)\iff -6\lt x\lt -\frac{28}{b}$$

Let $f(x)=x^2+(4+b)x+16$. Then, since the equation has only one solution, we have to have $$f(-6)f\left(-\frac{28}{b}\right)\lt 0\iff b\lt -14$$ So, in this case, $b\lt -14$.

Case 2 : $-14\leqslant b\leqslant -12$ or $4\leqslant b\lt \frac{14}{3}$

$$(2)\iff -6\lt x\lt 2$$

$b=4$ is sufficient, and $b=-12$ is not sufficient. For $b\not=4,-12$, $$f(-6)f(2)\lt 0\iff b\lt -14\quad\text{or}\quad b\gt \frac{14}{3}$$ So, in this case, $b=4$.

Case 3 : $b\geqslant \frac{14}{3}$

$$(2)\iff -\frac{28}{b}\lt x\lt 2$$ $b=\frac{14}{3}$ is sufficient. For $b\gt\frac{14}{3}$, $$f\left(-\frac{28}{b}\right)f(2)\lt 0\iff b\gt \frac{14}{3}$$ So, in this case, $b\geqslant 14/3$.

Therefore, the answer is $$\color{red}{(-\infty,-14)\cup{4}\cup\bigg[\frac{14}{3},\infty\bigg)}$$