3(square root)75 + 2(square root)27

I understand the number inside the square root should be divided into number that are perfect squares but as far as the number on the outside, 3 and 2, I am not too sure what to do with those.

Image will be attached to better see the problem.

Originally Posted by Eddyrodriguez
3(square root)75 + 2(square root)27

I understand the number inside the square root should be divided into number that are perfect squares but as far as the number on the outside, 3 and 2, I am not too sure what to do with those.

Image will be attached to better see the problem.
You have $3\sqrt{75} + 2\sqrt{21}$. Write $75=3\cdot 25$ and $27=3\cdot 9$. Factor out and square root the perfect squares and add like terms.

Originally Posted by Eddyrodriguez
3(square root)75 + 2(square root)27

I understand the number inside the square root should be divided into number that are perfect squares but as far as the number on the outside, 3 and 2, I am not too sure what to do with those.

Image will be attached to better see the problem.
The idea of "simplifying the radical expression", you want the smallest number inside the radical as possible. So, do what you suggested.

$a\sqrt{b^2c} = ab\sqrt{c}$

In your example, you have $75 = b^2c$ for some integers $b,c$. So, you have $3\sqrt{75} = 3\sqrt{b^2c} = 3b\sqrt{c}$

Okay so after writing out 75 = 3 • 25 and 27= 3 • 9 , the next step would be 75 = 3 (5 • 4) which would them become 3(5•2) after simplifying and 27 = 3(3) after simplifying correct?

Well, first, 25 is 5*5, not 5*4. And 27 is not "3(3)". Rather, $\displaystyle \sqrt{27}= \sqrt{9}\sqrt{3}= 3\sqrt{3}$ If you mean that $\displaystyle 3\sqrt{75}+ 2\sqrt{27}= 3\sqrt{5*25}+ 2sqrt{3*9}= 3\sqrt{5(5^2)}+ 2\sqrt{3(3^2)= 3(5\sqrt{5})+ 2(3\sqrt{3})= 15\sqrt{5}+ 6\sqrt{3}$ then that is correct.

No.$\sqrt{25\cdot 3} = \sqrt{25}\sqrt 3 = 5\sqrt 3$ and similarly for $27=3\cdot 9$. Then you can add them.