# How does one go about simplifying $\sqrt{72}$

In my book I am reading I sometimes see that the writer simplifies most of the answers most of the time. Take the following example.

I calculated an answer to the following $\sqrt{72}$, the book has the answer $6\sqrt{2}$.

Now these two answers are exactly the same. I would like to know how to get $\sqrt{72}$ to $6\sqrt{2}$, how does one calculate this? Is there a formula you use or a certain method which I am not aware of?

• $\sqrt{ab} = \sqrt{a}\sqrt{b}$ for $a>0, b>0$. – Braindead Dec 13 '14 at 14:46
• for $a\geq 0$ and $b\geq 0$ – Dr. Sonnhard Graubner Dec 13 '14 at 14:50

You factor $72=2^3\cdot 3^2$. Then you take the highest even power of each prime, so $72=(2\cdot 3)^2\cdot 2$ You can then pull out the square root of the product of the even powers. $\sqrt {72} = \sqrt{(2\cdot 3)^2\cdot 2}=(2\cdot 3)\sqrt 2=6 \sqrt 2$

We prime factorize $72=2^3 \cdot 3^2=2^2 \cdot 3^2 \cdot 2=(2\cdot 3)^2 \cdot 2$.

Hence $\sqrt{72}=\sqrt{(2\cdot 3)^2 \cdot 2}=\sqrt{6^2\cdot 2}=6\sqrt{2}$.

Not a formula, a procedure.

$\sqrt{72} = \sqrt{36\cdot 2} = \sqrt{36}\cdot\sqrt{2}=6\sqrt{2}$

Note: the formula $\sqrt{ab}=\sqrt{a}\sqrt{b}$ holds for all positive numbers $a$ and $b$. But remember never to use it for negative numbers (where the square root of a negative number would be imaginary) or you will run into nasty contradictions!

$\forall(a,b, \gamma)\in \mathbb{R}_+^2\times \mathbb{N}$: $$a^\gamma b^\gamma =(ab)^\gamma$$ So, $\sqrt{ab} = \sqrt{a}\sqrt{b}$, and here : $$\sqrt{72} = \sqrt{36\times2} = \sqrt{36}\sqrt{2} = 6\sqrt{2}$$