## anonymous 3 years ago Can someone help me with this radical?

1. anonymous

$\frac{ 7 }{ \sqrt{45} }$

2. anonymous

hmmm

3. anonymous

maybe before you do that, note that $$45=9\times 5$$ and so $$\sqrt{45}=\sqrt{9\times 5}=\sqrt{9}\sqrt{5}=3\sqrt{5}$$

4. anonymous

then $\frac{ 7 }{ \sqrt{45} }=\frac{7}{3\sqrt{5}}$and now you only need to multiply top and bottom by $$\sqrt{5}$$ to remove the radical from the denominator

5. anonymous

@julia_copen you got this or you need the steps?

6. anonymous

7. anonymous

lets start with $\frac{7}{3\sqrt{5}}$ your job is to get the radical out of the denominator, so multiply top and bottom by $$\sqrt 5$$ you get $\frac{7}{3\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}}{3\sqrt{5}\sqrt{5}}=\frac{7\sqrt{5}}{3\times 5}$

8. anonymous

typo there, meant $\frac{7}{3\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{7\sqrt{5}}{3\sqrt{5}\sqrt{5}}=\frac{7\sqrt{5}}{3\times 5}$

9. anonymous

final answer is $$\frac{7\sqrt{5}}{15}$$

10. anonymous

another quick example $\frac{4}{\sqrt{3}}=\frac{4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{4\sqrt{3}}{3}$

11. anonymous

Thanks so much! Me and my friend were having a hard time trying to understand how to break it down.

12. anonymous

yw

13. anonymous

can you help me with another?

14. anonymous

sure

15. anonymous

$\frac{ 1 }{ \sqrt{75z} }$

16. anonymous

is the $$z$$ inside the radical?

17. anonymous

yes

18. anonymous

ok the idea is to see if the number is the product of some "perfect square" so in this case $$75=25\times 3$$ making $\sqrt{75}=\sqrt{25}\sqrt{3}=5\sqrt{3}$ so star with $\frac{1}{5\sqrt{3z}}$ and then multiply top and bottom by $$\sqrt{3z}$$

19. anonymous

you get $\frac{1}{5\sqrt{3z}}\times \frac{\sqrt{3z}}{\sqrt{3z}}=\frac{\sqrt{3z}}{15z}$

20. anonymous

Okay I see. It's always difficult in the beginning for me.

21. anonymous

you will get used to it, (and then probably forget it because it is not really that useful) but in any case it gets easier

22. anonymous

Are you still on?