"Extraneous solution" solves original equation For the given equation:
$$x - 10 = \sqrt{9x}$$
when one simplifies, through the following steps:
\begin{align*}
x^2 - 20x + 100 &= 9x\\
x^2 - 29x + 100 &= 0\\
x &= 25, 4
\end{align*}
we check for extraneous solutions to make sure we have not altered from the set of solutions of the original equation while simplifying.
Hence $25$:
\begin{align*}
25-10 &= \sqrt{3 \times 3 \times 5 \times 5}\\
15 &= 15
\end{align*}
Then $4$:
\begin{align*}
4-10 &= \sqrt{36}\\
-6 &= \sqrt{36}
\end{align*}
Since $-6$ is a square root of $36$, 
$-6 = -6$. Solutions: $25$, $4$.
However, when I checked this through graphing, it appears that only $25$ is a solution and $4$ is considered extraneous. Why is this so? Square roots accounts for both a positive and a negative value and so the statement $-6 = \sqrt{36}$ should be true.
Looking at the graph of the equation, you could take the positive or the negative of the square root, ending up with two different graphs.
Why did we accept $25$ as a solution but reject $4$?
Any help/explanation would be much appreciated.

This is because you thought that $\sqrt{36}=\pm 6$ though it is not. Look when we do the checking at $x=4$, we get $x-10=4-10=-6$ whereas $\sqrt{9x}=\sqrt{36}=6$ which shows that $x-10=\sqrt{9x}$ does not hold when $x=4$. 
