SOLUTION: sqrt (56ab^3) / sqrt (7a)

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Question 201791: sqrt (56ab^3) / sqrt (7a)
Found 2 solutions by Theo, jsmallt9:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
= =
your problem becomes:

which becomes

because cancels out (same in numerator and denominator).
can be reduced further (taking most out from under square root sign that you can) because 8 is a product of and is a product of .
is 2 and is b, so the final reduced answer is

to prove your answer is correct, solve it both ways (with the original equation and the reduced equation).
let a = 10
let b = 20
with the reduced equation = 252.9822128
with the original equation = 252.9822128
since the answers are the same, the reduction is correct.
you can try some other numbers for a and b to prove it to yourself.
Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!
To simplify we will need a couple of basic properties of square roots:
and

We can use the first property to combine your fraction of square roots into a square root of a fraction:

We do this because we can cancel factors and reduce the fraction:

The 7's and the a's cancel leaving:

Now we factor out as many perfect squares as we can find:

We can use the Commutative property to rearrange the order:

And now we can use the second property to separate out the perfect square factors:

The .
And normally we would say that so that we can guarantee a non-negative value for the value of the square root. But if we look back at the original expression, we can see that:
  1. "a" cannot be zero because it would make the denominator zero.
  2. "a" cannot be negative because the radicand (the number inside the square root) must not be negative
  3. So "a" must be positive.
  4. Since "a" is positive must be zero or positive so that the radicand in the numerator is not negative.
  5. Since is zero or positive then "b" must be zero or positive.
  6. If "b" is zero or positive, then

So we do not need |b| to guarantee a non-negative square root. We can use just plain "b".
Substituting into our expression we get:

or simply
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