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Mathematics 17 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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..?

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

and how would you go about doing this?

OpenStudy (campbell_st):

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

OpenStudy (anonymous):

25 and h^2?

OpenStudy (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

OpenStudy (anonymous):

can you square root 5 though is that even possible?

OpenStudy (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 \[5 \times h^2 \times \sqrt{5}\]

OpenStudy (anonymous):

so wouldn't the answer be 5h^2√5?

OpenStudy (campbell_st):

just write it without the multiplication signs

OpenStudy (campbell_st):

that's correct

OpenStudy (phi):

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

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