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How do you simplify #sqrt6 - sqrt 3#?

Answer 1

Isaiah Anthony

#sqrt6-sqrt3=sqrt3(sqrt2-1)#

This is about as straightforward as it gets, though it can be factored if desired.

Recall that (for #a,b>0#)

#sqrt(ab)=sqrta*sqrtb#

Thus, we can state that

#sqrt6=sqrt(2*3)=sqrt2*sqrt3#

Next, we can reword the initial phrase:

#sqrt6-sqrt3=(sqrt2*sqrt3)-sqrt3#

Factor #sqrt3# from both terms:

#(sqrt2*sqrt3)-sqrt3=sqrt3(sqrt2-1)#

I wouldn't go so far as to say that #sqrt3(sqrt2-1)# is a simplification of #sqrt6-sqrt3#, but they are equivalent statements. Using #sqrt3(sqrt2-1)# may be helpful in simplifying more complicated expressions, such as:

#(sqrt6-sqrt3)/(sqrt10-sqrt5)=(sqrt3(sqrt2-1))/(sqrt5(sqrt2-1))=sqrt3/sqrt5=sqrt3/sqrt5(sqrt5/sqrt5)=sqrt15/5#

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Answer 2

Hazel Anglin

To simplify sqrt6 - sqrt3, we cannot combine the two terms since they have different radicals. Therefore, the expression remains as sqrt6 - sqrt3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.