How do you determine the correct way to simplify two radical numbers while dividing them and how do you know if the number remains squared or becomes a whole

4√63 / 6√36

Let us look at your examples:

4sqrt(63)=4sqrt(9 x 7). Notice that 9 is a perfect square. You can take its sqrt, which is 3 and multiply it by 4 outside the sqrt. So, it becomes: 3 x 4sqrt(7)=12sqrt(7), which is the same as: 4sqrt(63).

Now, the second example: 6sqrt(36)=6sqrt(6 x 6). Since 36 is a perfect square, you can take its sqrt, which 6. Then multiply this 6 by the 6 outside the sqrt, and you get: 6 x 6=36, which is the same as: 6sqrt(36). And that is it.

How do you determine the correct way to simplify two radical numbers while dividing them and how do I know if the number remains squared or becomes a whole number? Perhaps this should be used for an example? 4√63 / 6√36

$\begin{array}{rcll} \dfrac{ 4 \cdot \sqrt{63} } { 6 \cdot \sqrt{36} } &=& \dfrac{ 4 }{ 6 }\cdot \dfrac{ \sqrt{63} } { \sqrt{36} } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \dfrac{ \sqrt{63} } { \sqrt{36} } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \sqrt { \dfrac{ 63 } { 36 } } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \sqrt { \dfrac{ 9\cdot 7 } { 9\cdot 4 } } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \sqrt { \dfrac{ 7 } { 4 } } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \dfrac{ \sqrt{7} } { \sqrt{4} } \\\\ &=& \dfrac{ 2 }{ 3 }\cdot \dfrac{ \sqrt{7} } { 2 } \\\\ &=& \dfrac{ \sqrt{7} } { 3 } \\\\ \end{array}$