<s>[INST] 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...
 [/INST] $x = \frac 16$ 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 $
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