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. =$$-2/2\log_{5}(bx+28)=-\log_5(12-4x-x^2)$$ My try:
After removing the logarithmic terms I get the quadratic $x^2+x(b+4)+16=0$ Putting discriminant equal to $0$ I get $b={4,-12}$ But $-12$ cannot be a solution as it makes $12-4x-x^2$ negative so I get $b=4$ as the only solution.
But the answer given is $(-\infty,-14]\cup{4}\cup[14/3,\infty)$.I've no idea how. Help me please.
On
Find the values of 'b' for which the equation $$2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)\tag1$$ has only one solution.
After removing the logarithmic terms I get the quadratic $$x^2+x(b+4)+16=0\tag2$$
$$(1)\iff\ln(\text{b}x+28)=\ln(12-4x-x^2)\\\iff x^2+x(b+4)+16=0\quad\text{and}\quad bx+28>0\\\iff x=-\frac b2-2\pm\frac12\sqrt{(b+12)(b-4)}\quad\text{and}\quad x>-\frac{28}b.$$
Putting discriminant equal to $0$ I get $b={4,-12}$ But $-12$ cannot be a solution as it makes $12-4x-x^2$ negative so I get $b=4$ as the only solution.
You forgot to consider the case in which equation $(2)$ has two distinct roots exactly one of which satisfies $(1).$ From the above equivalences, this corresponds to $$-\frac b2-2-\frac12\sqrt{(b+12)(b-4)}\le -\frac{28}b\quad\text{and}\quad -\frac b2-2+\frac12\sqrt{(b+12)(b-4)}>-\frac{28}b,$$ whose solution, according to Desmos, is $$\left(-\infty,-14\right]\cup\left[\frac{14}3,\infty\right).$$
Combining the two cases gives $$\left(-\infty,-14\right]\cup\{4\}\cup\left[\frac{14}3,\infty\right),$$ as required.
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 $$\color{red}{(-\infty,-14)\cup{4}\cup\bigg[\frac{14}{3},\infty\bigg)}$$