document.write( "Question 363976: 1.explain why 3[sqrt] x^6=x2 for any value of x, but x[sqrt^6]= x3 only when x ≥ 0. \n" ); document.write( "

Algebra.Com's Answer #259669 by robertb(5830) $\"\"$ $\"About$

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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! \n" ); document.write( "

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