## julia_copen Group Title Can someone help me with this radical? one year ago one year ago

1. julia_copen Group Title

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

2. satellite73 Group Title

hmmm

3. satellite73 Group Title

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. satellite73 Group Title

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. satellite73 Group Title

@julia_copen you got this or you need the steps?

6. julia_copen Group Title

7. satellite73 Group Title

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. satellite73 Group Title

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. satellite73 Group Title

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

10. satellite73 Group Title

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

11. julia_copen Group Title

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

12. satellite73 Group Title

yw

13. julia_copen Group Title

can you help me with another?

14. satellite73 Group Title

sure

15. julia_copen Group Title

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

16. satellite73 Group Title

is the $$z$$ inside the radical?

17. julia_copen Group Title

yes

18. satellite73 Group Title

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. satellite73 Group Title

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

20. julia_copen Group Title

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

21. satellite73 Group Title

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. julia_copen Group Title

Are you still on?