Simplifying Square Roots Frustration Okay, I'm really frustrated with this.
So, when you have $3 \sqrt 5 + 5 \sqrt 5$,
you get $8\sqrt5$, right?
Okay, so what do I do for here:
$\sqrt{11} - 3 \sqrt{11}$
Is it just $-3  \sqrt{11}$ ?
What about for $6 \sqrt 2 + 4  \sqrt{50}$ ?
Do I multiply the $6 \sqrt{2}$ so that I can make what's inside of the square root equal to $50$?
 A: *

*$\sqrt{11}-3\sqrt{11}=\color{red}{1}\cdot\sqrt{11}-3\sqrt{11}=-2\sqrt{11}.$

*$6\sqrt 2+4\sqrt{50}=6\sqrt 2+4\color{blue}{\sqrt{2\cdot 5^2}}=6\sqrt{2}+4\cdot \color{blue}{5\sqrt 2}=6\sqrt 2+20\sqrt 2=26\sqrt 2$.
A: Sometimes it might help to take things a little more symbolically. Without worrying about what $x$ is for now, think about $3x + 5x$. Clearly that's $8x$, just as you thought.
For the next problem you have $x - 3x$. The answer is $-2x$. Let's say $x = 1$. Then this boils down to $1 - 3 = -2$. When we put $x = \sqrt{11}$ back into the picture, we get $\sqrt{11} - 3\sqrt{11} = -2\sqrt{11}$.
With the third problem, think about $x \neq y$. We have $6x + 4y$ and you're pretty much stuck unless you can figure out some way to restate $x$ in terms of $y$ or vice versa. It turns out that $\sqrt{50} = 5\sqrt{2}$ (so $y = 5x$). Therefore $$6\sqrt{2} + 4\sqrt{50} = 6\sqrt{2} + 4(5\sqrt{2}) = 6 + 20\sqrt{2} = 26\sqrt{2}.$$
A: Compare:
\begin{align}
  x - 3 x
  &= 1x - 3 x \\
  &= (1 - 3)x \\
  &= -2 x
\end{align}
to
\begin{align}
  \sqrt{11} - 3 \sqrt{11}
  &= 1\sqrt{11} - 3 \sqrt{11} \\
  &= (1 - 3)\sqrt{11} \\
  &= -2 \sqrt{11}
\end{align}
