# Simplifying square root

I need some help on simplifying square roots. I came across this example from my text which I'm not sure how they simplify

$$\sqrt {\frac {8}{9}} = \frac {2\sqrt{2}}{3}$$ and $$\sqrt {\frac {225}{4}} \cdot \frac {2\sqrt{2}}{3} = 5\sqrt{2}.$$

The LHS and RHS decimal value is equal based on my calculator. However, I'm not sure how they simply to example as above. Hope someone can enlighten me or introduce me a link to learn this which I believe is basic.

PS: My foundation of math is not strong therefore I need to pick up my basic before I can handle those tough questions. Thanks!

• A similar question is discussed here. Perhaps some of those answers can be of additional help to you, together with those below. – Cameron Buie Oct 22 '13 at 13:58

For the second problem, again we use the fact that $$\sqrt {ab} = \sqrt a \cdot \sqrt b$$ and we also use that fact that $$\sqrt{\frac ab} = \frac{\sqrt a}{\sqrt b}$$

So $$\sqrt{\dfrac{225}{4}} \cdot \frac{2 \sqrt 2}{3} = \dfrac{\sqrt {225}}{\sqrt 4} \cdot \frac {2\sqrt 2}{3} = \dfrac{\color{blue}{15}\cdot \color{red}{2} \sqrt 2}{\color{red}{2}\cdot \color{blue}3}= 5 \sqrt 2$$

• Hi amWhy, Thanks for the reply. Is there any website I can learn more about this with examples in case I miss out any other fact? – Withhelds Oct 22 '13 at 14:34
• See the Khan Academy for a lot of real help: video tutorials, practice problems, examples, etc. – Namaste Oct 22 '13 at 14:44

The rule that's being used here is $$\sqrt{xy} = \sqrt{x} \sqrt{y}$$

Hence $$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \sqrt{2} = 2\sqrt{2}$$

• Hi bluesh34, thanks for the reply. Can you explain on the * part (I have difficulty using bold :p $$\sqrt{8} = *\sqrt{4 \times 2} = \sqrt{4} \sqrt{2} = 2\sqrt{2}$$ * What if the number is e.g. $$\sqrt{31}$$ or other large number? – Withhelds Oct 22 '13 at 14:24
• $31$ is prime, so it cannot be factored. So in the case of $\sqrt {31}$, there is no simplification. – Namaste Oct 22 '13 at 14:36
• Thanks, I tried a few sum, i'm stuck for those odd factors like $$\sqrt{33} and \sqrt{96}$$ . – Withhelds Oct 22 '13 at 14:52
• When attempting to answer the question 'Can $\sqrt{x}$ be simplified, for a given integer $x$, the answer depends on whether or not $x$ is divisible by a square (number), such as $2 \times 2 = 4$ or $3 \times 3 = 9$. The squares up to 100 are 1,4,9,16,25,36,49,64,81,100. We don't need to check 1, since the fact that $\sqrt{x} = \sqrt{1 \times x} = \sqrt{1} \sqrt{x} = 1 \sqrt{x}$ doesn't simplify the expression. – George Tomlinson Oct 22 '13 at 19:02
• $\sqrt{33}$ isn't divisble by any of the other squares less than or equal to $16$ (we don't need to check any higher, as the only number greater than 16 = floor($\frac{33}{2}$) divisible by 33 is 33 itself), so it can't be simplified. (Floor(x) just means x - the decimal part of x). $\sqrt{96}$ is $\sqrt{4 \times 24} = \sqrt{4 \times 4 \times 6} = 2 \times 2 \times \sqrt{6} = 4 \sqrt6$. – George Tomlinson Oct 22 '13 at 19:03