Where is the mistake in solving the inequality? Where am I going wrong in solving this inequality?
$$\frac{p-\sqrt{9p-20}}{p-5}<2$$
On cross multiplying and squaring to remove the square root,I get the inequation
$p^2-29p+120<0$
Which gives $5<p<24$.
This is the correct answer http://www.wolframalpha.com/input/?i=solve+(p-root(9p-20))%2F(p-5)%3C2
But that is wrong answer and I can't understand why.Please help.
 A: First note each term is defined if $p\ge\dfrac{20}9$ and $p\ne 5$.
Now rewrite the inequation as 
\begin{align*}1+\frac{5-\sqrt{9p-20}}{p-5}<2&\iff\frac{5-\sqrt{9p-20}}{p-5}<1\iff \frac{45-9p}{(p-5)(5+\sqrt{9p-20})}<1\\
& \iff  \frac{-9}{(5+\sqrt{9p-20})}<1,
\end{align*}
which is always true since the l.h.s. is negative on the domain of the inequation:
$$\bigl[\tfrac{20}9,5\bigr)\cup (5,+\infty)$$
A: $$\frac{p-\sqrt{9p-20}}{p-5}<2$$
An inequality remains as it is only if you multiply both sides by a positive term. So, by multiplying by $p-5$, you have inherently assumed that $p-5>0$
Before multiplying, you have to specify a case for for the range of $p$ and proceed. Both cases need to be taken.
$$p-\sqrt{9p-20}<2p-10$$
$$-\sqrt{9p-20}<p-10$$
If $p-10<0$, then both the LHS and RHS would be negative and the inequality would change direction when squared. This results in $5<p<24$, of which $5<p<10$ falls within the assumptions taken on $p$.
If $p-10\ge0$, the RHS is non negative and the LHS is negative, which means the inequality is true for all $p\ge10$.
Finally, $p\ge\frac{20}{9}$, as the term in the square root must be positive.
Doing so should give the desired solution.
A: When you multiply an inequality by a negative number, the direction of the inequality is reversed.  Therefore, when we multiply by $p - 5$, we must consider two cases.  They are $p > 5$ or $p < 5$.  We do not need to consider $p = 5$ since the expression on the left hand side of the inequality 
$$\frac{p - \sqrt{9p - 20}}{p - 5} < 2$$
is undefined when $p = 5$.
Note that since we seek solutions in the real numbers, we require that 
$$9p - 20 \geq 0 \implies p \geq \frac{20}{9}$$
Case 1:  $p > 5$
If $p > 5$, then $p - 5 > 0$.  Hence, we can multiply both sides of the inequality by $p - 5$ without changing the direction of the inequality.
\begin{align*}
\frac{p - \sqrt{9p - 20}}{p - 5} & < 2\\
p - \sqrt{9p - 20} & < 2p - 10\\
10 - p & < \sqrt{9p - 20}
\end{align*}
If $p > 10$, then $10 - p$ is negative, while $\sqrt{9p - 20}$ is non-negative, so the inequality holds whenever $p > 10$.
If $5 < p \leq 10$, both sides of the inequality are non-negative, so we can square each side of the inequality without changing its direction.
\begin{align*}
100 - 20p + p^2 & < 9p - 20\\
p^2 - 29p + 120 & < 0\\
(p - 5)(p - 24) & < 0
\end{align*}
The inequality $(p - 5)(p - 24) < 0$ holds if $5 < p < 24$, which is the solution you obtained.  However, the restriction that $5 < p \leq 10$ means that the solution is restricted to $5 < p \leq 10$.  
Since the inequality holds if $5 < p \leq 10$ and if $p > 10$, we may conclude that it holds if $p > 5$ provided that $p \geq 20/9$.  Since $5 > 20/9$, the inequality is satisfied in this case if $p > 5$.  
Case 2:  $p < 5$
If $p < 5$, then $p - 5 < 0$.  Thus, multiplying each side of the inequality by $p - 5$ reverses the direction of the inequality.
\begin{align*}
\frac{p - \sqrt{9p - 20}}{p - 5} & < 2\\
p - \sqrt{9p - 20} & > 2p - 10\\
10 - p & > \sqrt{9p - 20}
\end{align*}
Since $p < 5$, both sides of the inequality are non-negative.  Therefore, the direction of the inequality is preserved if we square each side of the inequality.  Hence,
\begin{align*}
100 - 20p + p^2 & > 9p - 20\\
p^2 - 29p + 120 & > 0\\
(p - 5)(p - 24) & > 0
\end{align*}
The inequality $(p - 5)(p - 24)$ is valid if $p < 5$ or $p > 24$.  However, the restriction that $p < 5$ implies that the inequality holds if $p < 5$, provided that $p \geq 20/9$.  Thus, the inequality is satisfied in this case if $20/9 \leq p < 5$.  
Combining the results of cases 1 and 2 yields the solution set 
$$S = \left[\frac{20}{9}, 5\right) \cup (5, \infty)$$
