0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ $\sqrt{ab} = \sqrt{a}\sqrt{b}$ for $a>0, b>0$. $\endgroup$ – Braindead Dec 13 '14 at 14:46
  • $\begingroup$ for $a\geq 0$ and $b\geq 0$ $\endgroup$ – Dr. Sonnhard Graubner Dec 13 '14 at 14:50
6
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

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!

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.