SOLUTION: Simplify.{{{(2x)sqrt(2x)(3)sqrt(5x)}}} I need to check to verify this answer is correct: {{{(30x)sqrt(x)}}} Is this correct? Thanks.

Algebra ->  Square-cubic-other-roots -> SOLUTION: Simplify.{{{(2x)sqrt(2x)(3)sqrt(5x)}}} I need to check to verify this answer is correct: {{{(30x)sqrt(x)}}} Is this correct? Thanks.       Log On


   

Question 74516: Simplify.%282x%29sqrt%282x%29%283%29sqrt%285x%29
I need to check to verify this answer is correct: %2830x%29sqrt%28x%29 Is this correct? Thanks.
Found 2 solutions by Earlsdon, bucky:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Simplify:
2x%2Asqrt%282x%29%2A3%2Asqrt%285x%29 Rewrite as:
2x%2A3%2Asqrt%282x%29sqrt%285x%29+=+6x%2Asqrt%28%282x%29%2A%285x%29%29 = 6x%2Asqrt%282x%2A5x%29+=+6x%2Asqrt%2810x%5E2%29 = 6x%2Ax%2Asqrt%2810%29+=+6x%5E2%2Asqrt%2810%29

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Not correct.
.
This is a problem that might be easier to work in exponential form, using the rules of
exponents than it is to work it with the radical signs.
.
First, let's string out the problem to get:
.
2%2Ax%2Asqrt%282x%29%2A3%2Asqrt%285x%29
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Now let's assign exponents. Remember that the exponent %281%2F2%29 is equivalent to a radical
square root sign. After exponents are assigned the expression becomes:
.
%282%29%5E%281%29%2Ax%5E1%2A%282x%29%5E%281%2F2%29%2A+3%5E1+%2A+%285x%29%5E%281%2F2%29
.
Now recall the rule %28a%2Ab%29%5Ec=a%5Ec%2Ab%5Ec In other words, if two or more multiplied terms
are grouped in parentheses and an exponent applies to the entire group, then the exponent
can be applied to each of the multiplied terms.
.
In this problem the term %282x%29 is raised to the power %281%2F2%29. So the exponent
%281%2F2%29 can be applied individually to the 2 and to the x. The result is:
.
%282%2Ax%29%5E%281%2F2%29+=+2%5E%281%2F2%29+%2A+x%5E%281%2F2%29
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Substitute the right side of this into the expression and it becomes:
.

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Do the same kind of expansion on %285%2Ax%29%5E%281%2F2%29 to get 5%5E%281%2F2%29%2A+x%5E%281%2F2%29. Substitute
the right side of this into the expression and get:
.

.
Multiply all three of the x terms together by adding their exponents to get:
.
x%5E%281%29%2Ax%5E%281%2F2%29%2Ax%5E%281%2F2%29+=+x%5E%281%2B1%2F2%2B1%2F2%29+=+x%5E2
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Substitute this answer in as a replacement for the three x terms to get:
.
%282%29%5E%281%29%2A2%5E%281%2F2%29%2A3%5E1+%2A5%5E%281%2F2%29%2A+x%5E%282%29
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Convert the remaining terms back to radicals and you get:
.
2%2Asqrt%282%29%2A3%2Asqrt%285%29%2Ax%5E2
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Multiply sqrt%282%29 by sqrt%285%29 and the product is sqrt%2810%29
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Multiply 2 times 3 and the result is 6.
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So the three simplifications we got were x%5E2, sqrt%2810%29, and 6. Multiply these
three together to get the answer to the problem. The answer is:
.
6%2Asqrt%2810%29%2Ax%5E2
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Notice that the term sqrt%28x%29 does not appear in the answer at all.
.
Hope you can follow this and find why your answer is different from this one.