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) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt+%2856%2Aa%2Ab%5E3%29 = sqrt+%287%2Aa%2A8%2Ab%5E3%29 = sqrt%287%2Aa%29+%2A+sqrt+%288%2Ab%5E3%29
your problem becomes:
%28sqrt%287%2Aa%29+%2A+sqrt+%288%2Ab%5E3%29%29%2Fsqrt%287%2Aa%29
which becomes
sqrt%288%2Ab%5E3%29
because sqrt%287%2Aa%29 cancels out (same in numerator and denominator).
sqrt%288%2Ab%5E3%29 can be reduced further (taking most out from under square root sign that you can) because 8 is a product of 2%5E2%2A2 and b%5E3 is a product of b%5E2%2Ab.
sqrt%282%5E2%29 is 2 and sqrt%28b%5E2%29 is b, so the final reduced answer is
2%2Ab%2Asqrt%282%2Ab%29
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 2%2Ab%2Asqrt%282%2Ab%29 = 252.9822128
with the original equation sqrt%2856%2Aa%2Ab%5E3%29%2Fsqrt%287%2Aa%29 = 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) About Me  (Show Source):
You can put this solution on YOUR website!
To simplify sqrt+%2856ab%5E3%29+%2F+sqrt+%287a%29 we will need a couple of basic properties of square roots:
sqrt%28x%2Fy%29+=+sqrt%28x%29%2Fsqrt%28y%29 and
sqrt%28x%2Ay%29+=+sqrt%28x%29%2Asqrt%28y%29
We can use the first property to combine your fraction of square roots into a square root of a fraction:
sqrt+%2856ab%5E3%29+%2F+sqrt+%287a%29+=+sqrt%28%2856ab%5E3%29+%2F+%287a%29%29
We do this because we can cancel factors and reduce the fraction:
sqrt%28%2856ab%5E3%29+%2F+%287a%29%29+=+sqrt%28%287%2A8%2Aa%2Ab%5E3%29%2F%287a%29%29
The 7's and the a's cancel leaving:
sqrt%288b%5E3%29
Now we factor out as many perfect squares as we can find:
sqrt%284%2A2%2Ab%5E2%2Ab%29
We can use the Commutative property to rearrange the order:
sqrt%284%2Ab%5E2%2A2b%29
And now we can use the second property to separate out the perfect square factors:
sqrt%284%29%2Asqrt%28b%5E2%29%2Asqrt%282b%29
The sqrt%284%29+=+2.
And normally we would say that sqrt%28b%5E2%29+=+abs%28b%29 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 b%5E3 must be zero or positive so that the radicand in the numerator is not negative.
  5. Since b%5E3 is zero or positive then "b" must be zero or positive.
  6. If "b" is zero or positive, then abs%28b%29+=+b

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:
2%2Ab%2Asqrt%282b%29
or simply
2b%2Asqrt%282b%29