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I am having trouble simplifying a radical expression, such as say...$\sqrt{80}$.

What I do is firstly, I do 80/2, then 80/3, then 80/4, then 80/5...etc until I find the largest number that can be squared. It's very time consuming. It feels like I am doing something wrong. Can someone show me a quicker way to do this? I didn't really pay attention during class when we did these stuff.

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3 Answers 3

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What are the perfect squares under $80$?

$\sqrt{4}=2$, $\sqrt{9}=3$, $\sqrt{16}=4$, $\sqrt{25}=5$, $\sqrt{36}=6$, $\sqrt{49}=7$ and $\sqrt{64}=8$.

What is the largest radicand by which $80$ is divisible?

That will be $16$, so $\sqrt{80}=\sqrt{16}\sqrt{5}=4\sqrt{5}$.

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Knowing that $\sqrt{4}=2$ and in this case $80$ is divisible by $4$ you could do the following: $$\sqrt{80}=\sqrt{4 \cdot 20}=\sqrt{4} \sqrt{20}=2 \sqrt{4 \cdot 5}=2 \sqrt{4} \sqrt{5}=2 \cdot 2 \sqrt{5}=4 \sqrt{5}$$

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For this particular problem note that: $$80=10\times 8$$ $$80=(5\times2)\times(2\times4)$$ $$80=(5\times2)\times(2\times2\times2)$$ $$80=5\times2^4$$ so you have, $$\sqrt{80} =\sqrt{5\times2^4}=2^2\sqrt{5}=4\sqrt{5}$$ Generally you can try prime factorisation to get the simplest form, check out http://www.mesacc.edu/~scotz47781/mat120/notes/radicals/simplify/simplifying.html

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