Basic Square Root Question

**HenryJ** · Aug 1st 2009, 08:32 AM

I was reading an article on Purplemath about simplifying the square root terms. For instance in $\sqrt{144}$ they instruct users to break down as a product and simply from there $\sqrt{144}=\sqrt{9*16}=3*4=12$ .

However what if you used a different pair of factors? Such as $\sqrt{144}=\sqrt{3*48}$ ? I understand once you get a perfect square you can remove it from under the radicand, but I attempted this problem and could not figure it out with the pair of factors I have. $\sqrt{144}=\sqrt{3*48}=\sqrt{3}\sqrt{2x24}=\sqrt{3 }\sqrt{2}\sqrt{2x12}$ etc

I understand that we could simply solve by
knowing $12^2=144$ but that defeats the purpose of this exercise.

**VonNemo19** · Aug 1st 2009, 08:36 AM

Originally Posted by HenryJ

I was reading an article on Purplemath about simplifying the square root terms. For instance in $\sqrt{144}$ they instruct users to break down as a product and simply from there $\sqrt{144}=\sqrt{9*16}=3*4=12$ .

However what if you used a different pair of factors? Such as $\sqrt{144}=\sqrt{3*48}$ ? I understand once you get a perfect square you can remove it from under the radicand, but I attempted this problem and could not figure it out with the pair of factors I have. $\sqrt{144}=\sqrt{3*48}=\sqrt{3}\sqrt{2x24}=\sqrt{3 }\sqrt{2}\sqrt{2x12}$ etc

I understand that we could simply solve by
knowing $12^2=144$ but that defeats the purpose of this exercise.

$\sqrt{3}\sqrt{48}=\sqrt{3}\sqrt{3}\sqrt{16}=\sqrt{ 3}\sqrt{3}\sqrt{4}\sqrt{4}=(\sqrt{3})^2(\sqrt{4})^ 2$

**HenryJ** · Aug 1st 2009, 08:49 AM

Where did I go wrong?

$\sqrt{144}=\sqrt{3*48}=\sqrt{3}\sqrt{2*24}=\sqrt{3 }\sqrt{2}\sqrt{2*12}=\sqrt{3}\sqrt{2}\sqrt{2}\sqrt {3*4}$ $=(\sqrt{3})^2(\sqrt{2})^2*4=3*2*4=24$

**AlephZero** · Aug 1st 2009, 08:52 AM

Originally Posted by HenryJ

Where did I go wrong?

$\sqrt{144}=\sqrt{3*48}=\sqrt{3}\sqrt{2*24}=\sqrt{3 }\sqrt{2}\sqrt{2*12}=\sqrt{3}\sqrt{2}\sqrt{2}\sqrt {3*4}$ $=(\sqrt{3})^2(\sqrt{2})^2*4=3*2*4=24$

You have mistaken $\sqrt{4}$ for 4 in your 6th step.

**HenryJ** · Aug 1st 2009, 08:53 AM

So I have thank you!

**VonNemo19** · Aug 1st 2009, 08:55 AM

Originally Posted by HenryJ

So I have thank you!

The bottom line:

If you have factored properly, there will be no problems.

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