Solutions to $(4x^2+\frac{16}3x)^{\sqrt {3-x}}=(4x^2+\frac{16}3x)^{\sqrt {2x+11}-\sqrt{x+2}}$ $$(4x^2+\frac{16}3x)^{\sqrt {3-x}}=(4x^2+\frac{16}3x)^{\sqrt {2x+11}-\sqrt{x+2}}$$
I found the solutions to be $0, -\frac32, -1, -\frac43$
I can't figure out why any of those wouldn't work, but my textbook apparently only sees $-\frac43$ as valid...
Namely, the question asks for the sum of all the answers, and the correct answer is "$-\frac43$"
So am I wrong? If so, where? Am I missing a solution? Thanks.
Apparently $1/6$ is a solution as well... How do you get this? I still don't get their solution...
 A: A solution might arise because the base is $0$ and it is raised to a nonnegative power. Solving $4x^2+\frac{16}3x=0$ yields $x\in\{0,-4/3\}$. Both of these work, once you have checked that the make the exponents nonnegative real. 
So now assume $x$ is a solution that does not make the base $0$. $x$ might be such that it makes the base $\pm1$. We'll need to treat this separately as well for what comes next. Solving $4x^2+\frac{16}3x=1$ yields $x\in\{1/6,-3/2\}$. Both of these solutions are valid, since they make each side $1^{\text{real}}$.
Solving $4x^2+\frac{16}3x=-1$ yields $x=-\frac{2}{3}\pm\frac{\sqrt{7}}{6}$. These don't work out as solutions though, because at least on the left side, you have $(-1)^{\text{non-integer}}$.
OK, now ruling out any solutions $x$ that make the base $0$ or $\pm1$:
$$\begin{align}
(4x^2+\frac{16}3x)^{\sqrt {3-x}}&=(4x^2+\frac{16}3x)^{\sqrt {2x+11}-\sqrt{x+2}}\\
\implies \left|4x^2+\frac{16}3x\right|^{\sqrt {3-x}}&=\left|4x^2+\frac{16}3x\right|^{\sqrt {2x+11}-\sqrt{x+2}}\\
\end{align}$$
Applying the logarithm base $\left|4x^2+\frac{16}3x\right|$ is valid, since this base is positive but not $1$:
$$\begin{align}
{\sqrt {3-x}}&={\sqrt {2x+11}-\sqrt{x+2}}\\
\implies3-x&=2x+11-2\sqrt{(2x+11)(x+2)}+x+2\\
\implies-4x-10&=-2\sqrt{(2x+11)(x+2)}\\
\implies2x+5&=\sqrt{(2x+11)(x+2)}\\
\implies4x^2+20x+25&=2x^2+15x+22\\
\implies2x^2+5x+3&=0\\
\implies(2x+3)(x+1)&=0\\
\end{align}$$
So potentially, $-1$ and $-3/2$ are more solutions. Actually, we already established $-3/2$ is a solution. Some of the implications that led here were only one-way, so we should check $x=-1$. $x=-1$ makes the equation read $(-4/3)^2=(-4/3)^{3-1}$, which is true depending on how raising negative numbers to powers is defined. And this may be an issue. You are probably thinking that raising a negative number to the 2nd power just means multiply the base by itself. But generally, raising a negative number to a real power gives a nonreal result, and if you want raising to the 2nd power to give results that are "close" to the result from raising to the $1.9$th power, say, then you don't define raising to the 2nd power as doubled multiplication. For this reason, many computer systems behave this way and would define $(-4/3)^2$ as $\exp(2\ln(-4/3))$, where $\ln(-4/3)$ is defined using an (arbitrary) choice of branch of the logarithm that extends to the negative reals.
Because of the arbitrary choice that is made to do this, many other systems simply declare raising any negative number to any power to be undefined, even when the power is an integer. If that were the system in play, then $-1$ is not a valid solution, since both sides of the equation would be undefined. 
So either the sum of solutions is $0-\frac43+\frac16-\frac32=-\frac{8}{3}$ (if you don't count $-1$) or $0-\frac43+\frac16-\frac32-1=-\frac{11}{3}$ (if you do).
A: $x = \frac 16$ is one of the solutions to $ 4x^2-\frac {16}{3}x = 1$
Three ways to establish equality.
1) $4x^2 + \frac{16}{3} x = 0$ whose roots sum to $-\frac 43 $
2)  $4x^2 + \frac{16}{3} x = 1$ whose roots also sum to $-\frac 43 $
3) $\sqrt{3-x} = \sqrt{2x+11}- \sqrt{x+2}$ is equivalent to $x^2+\frac 52 x + \frac 32 = 0 $  whose roots sum to $-\frac 52 $
