anonymous one year ago Simplify the rational expression please help and explain with me?√125h^4 choices a.5h^2√5 b.25h^2√5 c.h√62.5 d.5√5h^4

1. anonymous

I have a idea what to do but the power of 4 makes me get lost,

2. campbell_st

look at the problem this way $\sqrt{5 \times 25 \times h^2 \times h^2}$ which terms can you take the square root of..?

3. anonymous

25 would be 5 and wouldn't h^2 and h^2 be crossed out? still new to this

4. campbell_st

well 25 to 5 is correct h^2 goes to h when you take the square root so the problem can be rewritten as $\sqrt{5} \times \sqrt{25} \times \sqrt{h^2} \times \sqrt{h^2} = \sqrt{5} \times 5 \times h \times h$ just simplify it

5. anonymous

you get the same as you had before= right?if to solve it

6. campbell_st

well it has been simplified now... all you need to do is tidy up the answer so it matches one of your choices

7. anonymous

and how would you go about doing this?

8. campbell_st

ok... in bits $\sqrt{5} \times 5 = 5 \sqrt{5}$ what does $h \times h =$

9. anonymous

25 and h^2?

10. campbell_st

h^2 is correct so its now $5 \times \sqrt{5} \times h^2 = 5 \times h^2 \times \sqrt{5}$ rememeber when muliplying, the order doesn't matter

11. anonymous

can you square root 5 though is that even possible?

12. campbell_st

ok... the solution in correct but in algebra will normally don't write the multiplication signs when writing a term... so how can you rewrite $5 \times h^2 \times \sqrt{5}$

13. anonymous

so wouldn't the answer be 5h^2√5?

14. campbell_st

just write it without the multiplication signs

15. campbell_st

that's correct

16. phi

$$\sqrt{5}$$ is theoretically possible... but it is not a "nice" number 2.236067977499789696409173668731276235440618359611525724270897245410520925637804899414414408378782275... it goes on forever (ugh!)