SOLUTION: This is 5 different problems I've been having problems with for a while. 4/sqrt5 4sqrtx^12/24sqrtx^12 15sqrt8x^16/5sqrt2x^4 (3sqrt5)(2sqrt10) (2sqrtx)(5sqrtx^3)

Algebra ->  Radicals -> SOLUTION: This is 5 different problems I've been having problems with for a while. 4/sqrt5 4sqrtx^12/24sqrtx^12 15sqrt8x^16/5sqrt2x^4 (3sqrt5)(2sqrt10) (2sqrtx)(5sqrtx^3)      Log On


   

Question 200938: This is 5 different problems I've been having problems with for a while.
4/sqrt5
4sqrtx^12/24sqrtx^12
15sqrt8x^16/5sqrt2x^4
(3sqrt5)(2sqrt10)
(2sqrtx)(5sqrtx^3)
Found 2 solutions by jim_thompson5910, jsmallt9:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first three to get you going in the right direction.


# 1

4%2Fsqrt%285%29 Start with the given expression.


%284%2Asqrt%285%29%29%2F%28sqrt%285%29%2Asqrt%285%29%29 Multiply both the numerator and denominator by sqrt%285%29


%284%2Asqrt%285%29%29%2F%28sqrt%285%2A5%29%29 Combine the roots.


%284%2Asqrt%285%29%29%2F%28sqrt%2825%29%29 Multiply


%284%2Asqrt%285%29%29%2F5 Take the square root of 25 to get 5



So 4%2Fsqrt%285%29=%284%2Asqrt%285%29%29%2F5




# 2


%284%2Asqrt%28x%5E12%29%29%2F%2824%2Asqrt%28x%5E12%29%29 Start with the given expression.


%284%2Asqrt%28x%5E12%29%29%2F%284%2A6%2Asqrt%28x%5E12%29%29 Factor 24 to get 4*6


%28cross%284%29%2Asqrt%28x%5E12%29%29%2F%28cross%284%29%2A6%2Asqrt%28x%5E12%29%29 Cancel out the common terms.


%28sqrt%28x%5E12%29%29%2F%286%2Asqrt%28x%5E12%29%29 Simplify


sqrt%28%28x%5E12%29%2F%28x%5E12%29%29%2F%286%29 Combine the roots


sqrt%281%29%2F6 Divide


1%2F6 Simplify


So %284%2Asqrt%28x%5E12%29%29%2F%2824%2Asqrt%28x%5E12%29%29=1%2F6 where x%3E0




# 3


%2815%2Asqrt%288x%5E16%29%29%2F%285%2Asqrt%282x%5E4%29%29 Start with the given expression.


%285%2A3%2Asqrt%288x%5E16%29%29%2F%285%2Asqrt%282x%5E4%29%29 Factor 15 into 5*3


%28cross%285%29%2A3%2Asqrt%288x%5E16%29%29%2F%28cross%285%29%2Asqrt%282x%5E4%29%29 Cancel out the common terms.


%283%2Asqrt%288x%5E16%29%29%2F%28sqrt%282x%5E4%29%29 Simplify


3%2Asqrt%28%288x%5E16%29%2F%282x%5E4%29%29 Combine the roots.


3%2Asqrt%28%284%2A2x%5E16%29%2F%282x%5E4%29%29 Factor 8 into 4*2


3%2Asqrt%28%284%2Across%282%29x%5E16%29%2F%28cross%282%29x%5E4%29%29 Cancel out the common terms.


3%2Asqrt%28%284x%5E16%29%2F%28x%5E4%29%29 Simplify


3%2Asqrt%284x%5E%2816-4%29%29 Divide the variable terms by subtracting the exponents.


3%2Asqrt%284x%5E12%29 Subtract


3%2Asqrt%284%2Ax%5E2%2Ax%5E2%2Ax%5E2%2Ax%5E2%2Ax%5E2%2Ax%5E2%29 Factor x%5E12 into x%5E2%2Ax%5E2%2Ax%5E2%2Ax%5E2%2Ax%5E2%2Ax%5E2 note: there are 6 x%5E2 terms.


Break up the square root using the identity sqrt%28A%2AB%29=sqrt%28A%29%2Asqrt%28B%29.


Take the square root of 4 to get 2.


3%2A2%2Ax%2Ax%2Ax%2Ax%2Ax%2Ax Take the square root of x%5E2 to get x.


6x%5E6 Multiply.

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Answer:


So %2815%2Asqrt%288x%5E16%29%29%2F%285%2Asqrt%282x%5E4%29%29 simplifies to 6x%5E6


In other words, %2815%2Asqrt%288x%5E16%29%29%2F%285%2Asqrt%282x%5E4%29%29=6x%5E6 where x%3E0



Edit: The other poster is correct in saying that "x" doesn't need to be non-negative, BUT recall that sqrt%28x%5E2%29=abs%28x%29. If we assume that x%3E=0 (ie it is NOT negative), then sqrt%28x%5E2%29=x which simplifies things greatly. If "x" can be any number, then you must include the absolute value.

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I just want to correct something in the solutions of jimthompson5910:

In problems #2 and #3, x does not have to be positive. x could be positive or negative. The only number x cannot be is zero.