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

The difficulty is that your solution to the original inequality is wrong. There is no reason to say $\displaystyle x\ne \frac{1}{3}$. It is $\displaystyle x= \frac{1}{3}$ that would cause the left side to not exist.

sorry that was a typo, so why shouldn't the solution include $\displaystyle \sqrt{x}\neq\frac{1}{3}$
$\displaystyle x\neq \frac{1}{9}$

$\displaystyle \sqrt{x}$ is never negative so there no need to exclude 1/3. And, if x= 1/9, then [tex]\sqrt{x}= 1/3[/itex], not 1/3.

Problem 1
$\displaystyle \frac{(x+1)(4x)}{(3x+1)^2} \geq 0$
Let's change the equation up a bit to really look at it. All we'll do is pull a 1 out of the $\displaystyle (4x)$ to give $\displaystyle (x4)$, then rearrange it:
$\displaystyle \frac{(x+1)(x4)}{(3x+1)^2} \geq 0$
So this is what we already know by looking at it:
1. We have a negative degree two polynomial expression on top
2. This expression has two real 0s so at some point x will be > 0
2. We can easily figure out our zeros now for the equation (1 and 4)
3. Also it looks like if $\displaystyle x = \frac{1}{3}$ because we cannot divide by 0, there would be no solution.
So our answer is:
$\displaystyle 1 \leq x \leq 4$ and $\displaystyle x \neq \frac{1}{3}$
Note another way to write this solution:
$\displaystyle 1 \leq x < \frac{1}{3}$ and $\displaystyle \frac{1}{3} < x \leq 4$
Problem 2
Now on your next problem:
$\displaystyle \frac{(\sqrt{x}+1)(4\sqrt{x})}{(3\sqrt{x}+1)^2} \geq 0$
$\displaystyle \frac{(\sqrt{x}+1)(\sqrt{x}4)}{(3\sqrt{x}+1)^2} \geq 0$
It's going to be mostly the same except for two differences:
1. Solving for our roots won't be as easy, for example:
$\displaystyle (\sqrt{x}  4) = 0$
$\displaystyle \sqrt{x} = 4$
$\displaystyle x = 16$
$\displaystyle (\sqrt{x}+1) = 0$
$\displaystyle \sqrt{x} = 1$
There is no solution here, the sqrt(x) will never be negative.
No solution, so only 16 for a root.
So that was a little different but not too bad.
2. There will never be a real value for x that would create a 0 in the bottom of your expression on the right
For real numbers in x: $\displaystyle (3\sqrt{x}+1)^2 \neq 0$
So we don't have to worry about a "no solution" problem on the bottom.
And we understand that sqrt{x} will always remain nonnegative. To explain this better, if we put a negative number in for x, we get a complex number in return. If we put a 0 in for x, we get a 0. And if we put a positive number in for x, we get a positive in return. Never a negative result.
We can form our answer now to be $\displaystyle \sqrt{x} \leq 4$
Or in other words, because $\displaystyle \sqrt{x}$ is nonnegative: $\displaystyle 0 \leq \sqrt{x} \leq 4$
And thusly we have our solution: $\displaystyle 0 \leq x \leq 16$
Good luck you were on the right track it just looks like some typos threw you off.

$\displaystyle \displaystyle \frac{(x + 1)(4  x)}{(3x + 1)^2} \geq 0$
You first need to note that $\displaystyle \displaystyle x \neq  \frac{1}{3}$, because that would give a zero denominator.
Since the denominator is always nonnegative, you can multiply both sides by the denominator without changing the inequality sign
$\displaystyle \displaystyle (x + 1)(4  x) \geq 0$
Quadratic inequalities are nearly always easiest solved by completing the square.
$\displaystyle \displaystyle \begin{align*} (x + 1)(4  x) &\geq 0 \\ x^2 + 3x + 4 &\geq 0 \\ (x^2  3x  4) &\geq 0 \\ \left[x^2  3x + \left(\frac{3}{2}\right)^2  \left(\frac{3}{2}\right)^2  4\right] &\geq 0 \\ \left[\left(x  \frac{3}{2}\right)^2  \frac{9}{4}  \frac{16}{4}\right] &\geq 0 \\ \left[\left(x  \frac{3}{2}\right)^2  \frac{25}{4}\right] &\geq 0 \\ \left(x  \frac{3}{2}\right)^2 + \frac{25}{4} &\geq 0 \\ \left(x  \frac{3}{2}\right)^2 &\geq \frac{25}{4} \\ \left(x  \frac{3}{2}\right)^2 &\leq \frac{25}{4} \\ \leftx  \frac{3}{2}\right^2 &\leq \frac{5}{2} \\ \frac{5}{2} \leq x  \frac{3}{2} &\leq \frac{5}{2} \\ 1 \leq x &\leq 4\end{align*}$
Now, remembering that $\displaystyle \displaystyle x \neq \frac{1}{3}$, that means our solution is
$\displaystyle \displaystyle x \in \left[1, \frac{1}{3}\right) \cup \left(\frac{1}{3}, 4\right]$

Alternatively,
the fraction is undefined for $\displaystyle x=\frac{1}{3}$
For all other x, the denominator is positive.
Then for $\displaystyle x\ne\ \frac{1}{3}$
the solution reduces to $\displaystyle (x+1)(4x)\ge \0$
which will be true if both factors are negative or both are positive, or either is zero.
Both are negative if $\displaystyle x< 1$ and $\displaystyle x>4$ at the same time, which is not possible.
Both are positive if $\displaystyle 1< x< 4$
Therefore, both factors cannot be negative and so for $\displaystyle x\ne\ \frac{1}{3}$
$\displaystyle 1\le\ x\le\ 4$