# Thread: Basic Square Root Question

1. ## Basic Square Root Question

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

However what if you used a different pair of factors? Such as $\displaystyle \sqrt{144}=\sqrt{3*48}$? 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. $\displaystyle \sqrt{144}=\sqrt{3*48}=\sqrt{3}\sqrt{2x24}=\sqrt{3 }\sqrt{2}\sqrt{2x12}$ etc

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

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

However what if you used a different pair of factors? Such as $\displaystyle \sqrt{144}=\sqrt{3*48}$? 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. $\displaystyle \sqrt{144}=\sqrt{3*48}=\sqrt{3}\sqrt{2x24}=\sqrt{3 }\sqrt{2}\sqrt{2x12}$ etc

I understand that we could simply solve by
knowing $\displaystyle 12^2=144$ but that defeats the purpose of this exercise.
$\displaystyle \sqrt{3}\sqrt{48}=\sqrt{3}\sqrt{3}\sqrt{16}=\sqrt{ 3}\sqrt{3}\sqrt{4}\sqrt{4}=(\sqrt{3})^2(\sqrt{4})^ 2$

3. Where did I go wrong?

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

4. Originally Posted by HenryJ
Where did I go wrong?

$\displaystyle \sqrt{144}=\sqrt{3*48}=\sqrt{3}\sqrt{2*24}=\sqrt{3 }\sqrt{2}\sqrt{2*12}=\sqrt{3}\sqrt{2}\sqrt{2}\sqrt {3*4}$ $\displaystyle =(\sqrt{3})^2(\sqrt{2})^2*4=3*2*4=24$
You have mistaken $\displaystyle \sqrt{4}$ for 4 in your 6th step.

5. So I have thank you!

6. Originally Posted by HenryJ
So I have thank you!
The bottom line:

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