Simplifying Square Roots Frustration

Question

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 $50$?

Answer 1

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

• $6\sqrt 2+4\sqrt{50}=6\sqrt 2+4\color{blue}{\sqrt{2\cdot 5^2}}=6\sqrt{2}+4\cdot \color{blue}{5\sqrt 2}=6\sqrt 2+20\sqrt 2=26\sqrt 2$.

Answer 2

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 $$6\sqrt{2} + 4\sqrt{50} = 6\sqrt{2} + 4(5\sqrt{2}) = 6 + 20\sqrt{2} = 26\sqrt{2}.$$

Answer 3

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}