Fractions with radicals in the denominator I'm working my

Polchinio3es

Polchinio3es

Answered question

2022-03-24

Fractions with radicals in the denominator
I'm working my way through the videos on the Khan Academy, and have a hit a road block. I can't understand why the following is true:
668585=85

Answer & Explanation

Riya Erickson

Riya Erickson

Beginner2022-03-25Added 12 answers

With respect to handling square roots in the denominator:
Say, for example, you have ab
If you want a "square-root free" denominator, remember that you can multiply any fraction by the number 1 (or its equivalent) without changing its value. In this case, we want to multiply the given fraction by bb=1. Can you see why we would want to choose this particular representation of 1?
Now,
bbab=ab(b)2=abb
And if the greatest common divisor of a and b > 1, then we can reduce ab by dividing each of a and b by their greatest common divisor (i.e., reducing the fraction ab
Now, we take a look at a slightly more complicated situation, as in your problem, where the denominator is itself a fraction given by 668585
We can first multiply numerator and denominator by 85, then proceed as we did in the case above:
6685858585=685685=8585=85858585=858585=85
More simply, once we have reduced the expression to 8585, we notice that 85=(85)2, and so, this, together with the fact that we can then cancel a common factor of both the numerator and the denominator, we have:
8585=(85)285=85
Note: When you have enough experience simplifying fractions and fractions with fractions with roots, etc., many of the steps which I've displayed explicitly above will be "second nature" for you, and you'll be able to manipulate such expressions with far less in the way of computations than I've shown above.
Jamie Maldonado

Jamie Maldonado

Beginner2022-03-26Added 10 answers

No one seems to have posted the really simple way to do this yet:
8585=858585
and then cancel the common factor.

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