First part of the question says: Solve the inequality $\displaystyle \frac{(x+1)(4-x)}{(3x+1)^2}\geqslant 0$

And after solving, the answer is $\displaystyle -1\leqslant x\leqslant 4, x\neq \frac{1}{3}$

Then the next part of the question says: Hence deduce the solution to the inequality $\displaystyle \frac{(\sqrt{x}+1)(4-\sqrt{x})}{(3\sqrt{x}+1)^2}\geqslant 0$

On comparing, i replaced x in the answer to the first part with $\displaystyle \sqrt{x}$

so, $\displaystyle -1\leqslant \sqrt{x}\leqslant 4, \sqrt{x}\neq -\frac{1}{3}$

$\displaystyle -1\leqslant \sqrt{x}$ and $\displaystyle \sqrt{x}\leqslant 4$

$\displaystyle x\geqslant 0$ and $\displaystyle x\leqslant 16$

therefore, $\displaystyle 0\leqslant x\leqslant 16$

Yes, that's right. The answer is correct but shouldn't the answer $\displaystyle x\neq -\frac{1}{3}$ also be dealt with?

So shouldn't the right answer be... $\displaystyle 0\leqslant x\leqslant 16, x\neq \frac{1}{9}$ ? why is this not the answer? Please advise, thanks