SOLUTION: I have to problems i have been trying to solve please let me know if i am right or wrong. Use property 1 to simplify assume that all variables represent positive real numbers. 1

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Question 77916: I have to problems i have been trying to solve please let me know if i am right or wrong. Use property 1 to simplify assume that all variables represent positive real numbers.
1. sqrt50
= sqrt10%2A5
= sqrt10*5
= 5sqrt5
2. sqrt72x%5E3
=9*x^2*8x
=3xsqrt8x=3xsqrt2x
Found 2 solutions by Earlsdon, bucky:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Simplify:
1)
sqrt%2850%29+=+sqrt%2825%2A2%29=5%2Asqrt%282%29
2)
sqrt%2872x%5E3%29+=+sqrt%2836x%5E2%2A2x%29=6x%2Asqrt%282x%29

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
.
1. sqrt%2850%29
= sqrt%2810%2A5%29 <---- should be sqrt%2825%2A2%29
= sqrt%2810%29%2A%285%29 <--- should be sqrt%2825%29%2A+sqrt%282%29
= 5%2Asqrt%285%29 <--- should be 5%2Asqrt%282%29
The thing to do is to begin by looking at 50 and trying to find some factor or factors
that are squares. In this case 25 is a factor of 50 that has a "nice" square root of 5.
The comments after the <---- show you the way to simplify the problem step by step and the
answer to this problem is 5%2Asqrt%282%29
.
2. sqrt%2872x%5E3%29
=9%2Ax%5E2%2A8x<---- the biggest factor of 72 that is a square is 36. Use 36x%5E2%2A2x
This converts the problem to sqrt%2836%2Ax%5E2%29%2Asqrt%282%2Ax%29
.
=3x%2Asqrt%288x%29+=3x%2Asqrt%282x%29 <---- should be 6x%2Asqrt%282x%29
.
You had the right idea here. Your last step almost got the correct answer. Your last
step should have been:
.
=
.
and this is the correct answer.
.
Hope this gives you a little more insight to the problem. You seem to have the basic idea
of solving problems such as these and just needed a couple of minor suggestions to get you
back on track. Keep up the good work.