Square root of (4x^6)

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Square root of (4x^6)

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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sqrt(4) * x^(6/2)
No, not zero.
Simplify what amistre has written

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Other answers:

Why did he divide the 6 by 2..?
those root radicals are just exponents
2rt() = ^1/2 10rt() = ^1/10
because \[ (x^a)^b = x^{ab} \] Hence \[ \sqrt{x^6} = (x^6)^{1/2} = x^{6/2} = ... \]
its makes the math simpler
Hence here \[ \sqrt{4x^6} = \sqrt{4} \sqrt{x^6} = ... what? \]
I still don't get where the 1/2 is coming from
\[ \sqrt{x} = x^{1/2} \] For example, if \[ \sqrt{x} = 3 \] then \[ \sqrt{x}^2 = 3^2 \] i.e., \[ x = 3^2 \] It must be therefore that \( \sqrt{x} = x^{1/2} \). If this weren't the case we wouldn't have \[ \sqrt{x}^2 = x \] Having \[ \sqrt{x} = x^{1/2} \] is consistent with the definition of square root because \[ (x^{1/2})^2 = x^{2/2} = x^1 = x \]
So am i only dividing the x^6 by two to make it x^3 since the exponent doesn't pertain to the 4?

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