SOLUTION: 1.explain why 3[sqrt] x^6=x2 for any value of x, but x[sqrt^6]= x3 only when x ≥ 0.

Algebra ->  Radicals -> SOLUTION: 1.explain why 3[sqrt] x^6=x2 for any value of x, but x[sqrt^6]= x3 only when x ≥ 0.       Log On


   

Question 363976: 1.explain why 3[sqrt] x^6=x2 for any value of x, but x[sqrt^6]= x3 only when x ≥ 0.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The cube root of x%5E6 is equal to x%5E2 for any value of x because the sign of x is retained by the presence of the power 2. But sqrt%28x%5E6%29 can take on only non-negative values (because it is a square root!). Technically speaking, sqrt%28x%5E6%29 = |x%5E3|, and so there would be a problem if x<0. For example, if we allow x <0, then sqrt%28%28-2%29%5E6%29+=+%28-2%29%5E3+=+-8, but a square root is always non-negative!