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Mathematics 19 Online
OpenStudy (anonymous):

What is the square root of x^2+4x+4?

OpenStudy (anonymous):

\[\sqrt{ x^2+4x+4}\]

OpenStudy (anonymous):

exactly

OpenStudy (anonymous):

x+2

OpenStudy (anonymous):

Technically correct, the best kind of correct.

OpenStudy (anonymous):

+or -

OpenStudy (anonymous):

So there is no way to simplify?

OpenStudy (anonymous):

There is, honeyboyseab is correct. It is plus or minus x+2

OpenStudy (anonymous):

x^2+4x+4 factors to (x+2)^2

OpenStudy (anonymous):

yes the terms under the square root sign is a perfect square and resolves to (x+2)^2 and the square root of that is (x+2)

OpenStudy (anonymous):

Ah, I see now. Thanks!

OpenStudy (anonymous):

we have to break this function as...x^2+4x+4 we can write it like..x^2+2x+2x+4 now, we can take some common term here x(x+2)+2(x+2) Now again take common term here (x+2)(x+2)=(x+2)^2

OpenStudy (anonymous):

and sqrt of(x+2)^2=x+2

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