How do you simplify fractions with square roots?

Answer 1

See below:

Let's say I have:

#5/sqrt2#

How do I simplify this?

Remember that #sqrt2xxsqrt2=2# And so if I can multiply the denominator by #sqrt2#, we'll get 2 for the denominator. And that can be done, so long as we multiply both the numerator and denominator by the same thing (which means we're multiplying the fraction by 1 and not changing its value):
#5/sqrt2(1)=5/sqrt2(sqrt2/sqrt2)=(5sqrt2)/2#

I can do the same type of thing if I have a mixed radical in my denominator. Let's say it's:

#5/(3-sqrt2)#
If I multiply two types of this kind of number, one adding the square root and the other subtracting, I can simplify. Let's look at our example again. With #3-sqrt2#, we can multiply by #3+sqrt2#:
#(3-sqrt2)(3+sqrt2)=9-3sqrt2+3sqrt2-2=7#

And so for the full fraction, we have:

#5/(3-sqrt2)((3+sqrt2)/(3+sqrt2))=(15+5sqrt2)/7#
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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.

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.

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.

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.

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