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Mathematics 7 Online
OpenStudy (anonymous):

1. Please simplify sqrt(x^7).

OpenStudy (anonymous):

\[\sqrt{x^7} =\sqrt{x^2*x^2*x^2*x}\] can you tell me what the \[\sqrt{x^2} \]is?

OpenStudy (anonymous):

Can you please explain this to me?

OpenStudy (anonymous):

x

OpenStudy (anonymous):

well ok do you know what the square root of 4 is? \[\sqrt{4}\]

OpenStudy (anonymous):

2

OpenStudy (anonymous):

Its 2

OpenStudy (anonymous):

all you have do to is ask yourself what two number of the same multiply together give me 4 in this case 2 since 2x2=4

OpenStudy (anonymous):

But how would you do that if the number is 7?

OpenStudy (anonymous):

Do you know why the square root of 4 is 2?

OpenStudy (anonymous):

there a rule for numbers like 7

OpenStudy (anonymous):

Whats the rule?

OpenStudy (anonymous):

beside 7 isnt a perfect squirt

OpenStudy (anonymous):

I know this

OpenStudy (anonymous):

the sqrt of 7 is 7 since 7 isnt a perfect quare

OpenStudy (anonymous):

So the answer is 7 to my questions?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

That's not the answer.

OpenStudy (anonymous):

Then someone please explain to me step by step how to get the answer.

OpenStudy (anonymous):

\[\sqrt{x^7}=\sqrt{x^2*x^2*x^2*x}=\sqrt{x^2}*\sqrt{x^2}*\sqrt{x^2}*\sqrt{x}\]

OpenStudy (anonymous):

Sqrt(x*x*x*x*x*x*x) So, each x has to has a pair to leave the "square root" So, x-x x-x x-x x So there are three pairs and one left over SO! x^3 sqrt(x) :) hope it helped!

OpenStudy (anonymous):

Can you tell me what \[\sqrt{x^2}\]

OpenStudy (anonymous):

yeap just x

OpenStudy (anonymous):

I know you know I want to the Original asker to tell me what it is

OpenStudy (anonymous):

because when you multiply back x and x you get x sq

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