SOLUTION: Can anyone help me with this? My problem is how to rationalize the denominator: 2/ sqrt[6] - sqrt[5]

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Question 74896: Can anyone help me with this?
My problem is how to rationalize the denominator:
2/ sqrt[6] - sqrt[5]

Found 2 solutions by ptaylor, bucky:
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
SEEMS LIKE I'VE DONE THIS ONE BEFORE.

2%2F%28sqrt%286%29-sqrt%285%29%29 First, multiply both numerator and denominator by %28sqrt%286%29%2Bsqrt%285%29%29 we can do this because %28sqrt%286%29%2Bsqrt%285%29%29%2F%28sqrt%286%29%2Bsqrt%285%29%29 equals 1 and we want to do this to get rid of the radicals in the denominator.
and this equals

%28%282%29%28sqrt%286%29%2Bsqrt%285%29%29%29%2F%286-5%29 and this reduces to:
2%28sqrt%286%29%2Bsqrt%285%29%29 ------------ans

Hope this helps----ptaylor


Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
2%2F+%28sqrt%286%29+-+sqrt%285%29%29
.
I'm going to assume that both the square root terms are in the denominator as shown above.
.
You can rationalize the denominator through using the identity:
.
%28a+%2B+b%29%2A%28a+-+b%29+=+a%5E2+-+b%5E2
.
You can multiply out the left side of this identity just to convince yourself that the result
equals the right side.
.
The denominator of the problem is of the form (a - b). Suppose that we multiply the entire
term of the problem by %28a+%2B+b%29%2F%28a+%2B+b%29 which is equivalent to multiplying the term
in the problem by 1 since the numerator of this multiplier equals the denominator.
I used "a" and "b" so it might be easier to see what we're trying to do. Actually, we're
going to let:
.
a+=+sqrt%286%29 and b+=+sqrt%285%29.
.
So when we multiply the original problem by %28a+%2B+b%29%2F%28a+%2B+b%29 we're actually going
to multiply it by %28sqrt%286%29%2Bsqrt%285%29%29%2F%28sqrt%286%29%2Bsqrt%285%29%29.
.
Let's do it:
.

.
In accordance with the identity above, after multiplying the original denominator
of the problem by %28sqrt%286%29%2Bsqrt%285%29%29 the new, rationalized denominator becomes
the difference between the squares of the two terms that were in the original denominator:
.

.
But %28sqrt%286%29%29%5E2+=+6 and %28sqrt%285%29%29%5E2+=+5 so the new, rationalized denominator
just equals 6 - 5 or 1.
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Now all we have left to do is to work on the numerator.
.
The numerator of the original problem is 2, but we multiplied it by %28sqrt%286%29%2Bsqrt%285%29%29
as part of the process associated with rationalizing the denominator. When we do this we get:
.
2%2A%28sqrt%286%29%2Bsqrt%285%29%29+=+2%2Asqrt%286%29+%2B+2%2Asqrt%285%29
.
That's the answer ... 2%2Asqrt%286%29%2B2%2Asqrt%285%29 because the new denominator that it is over
is just 1.
.
Hope that after looking through this you'll be familiar with using the identity to rationalize
denominators. It will be useful to you when you work with complex numbers.
.