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anonymous

  • 5 years ago

Square root of (4x^6)

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  1. amistre64
    • 5 years ago
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    sqrt(4) * x^(6/2)

  2. JamesJ
    • 5 years ago
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    No, not zero.

  3. JamesJ
    • 5 years ago
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    Simplify what amistre has written

  4. anonymous
    • 5 years ago
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    Why did he divide the 6 by 2..?

  5. amistre64
    • 5 years ago
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    those root radicals are just exponents

  6. amistre64
    • 5 years ago
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    2rt() = ^1/2 10rt() = ^1/10

  7. JamesJ
    • 5 years ago
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    because \[ (x^a)^b = x^{ab} \] Hence \[ \sqrt{x^6} = (x^6)^{1/2} = x^{6/2} = ... \]

  8. amistre64
    • 5 years ago
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    its makes the math simpler

  9. JamesJ
    • 5 years ago
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    Hence here \[ \sqrt{4x^6} = \sqrt{4} \sqrt{x^6} = ... what? \]

  10. anonymous
    • 5 years ago
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    I still don't get where the 1/2 is coming from

  11. JamesJ
    • 5 years ago
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    \[ \sqrt{x} = x^{1/2} \] 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 \[ (x^{1/2})^2 = x^{2/2} = x^1 = x \]

  12. anonymous
    • 5 years ago
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    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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