Which of the following radical expressions has an absolute value symbol in its simplified form?

root(32x ^ 5, 5)
root(64x ^ 6, 6)
root(16x, 4)
root(- 8x ^ 3, 3)

1 answer

To determine which of the given radical expressions has an absolute value symbol in its simplified form, we need to analyze the roots and their properties.

  1. \(\sqrt[5]{32x^5}\):

    • The fifth root of \(32x^5\) can be simplified. Since 32 is \(2^5\), we have: \[ \sqrt[5]{32x^5} = \sqrt[5]{2^5 \cdot x^5} = 2x. \]
    • No absolute value since \(x^5\) retains its sign for real \(x\).
  2. \(\sqrt[6]{64x^6}\):

    • The sixth root of \(64x^6\) simplifies as follows. Since 64 is \(2^6\), we have: \[ \sqrt[6]{64x^6} = \sqrt[6]{2^6 \cdot x^6} = 2x. \]
    • Again, no absolute value since \(x^6\) retains its sign.
  3. \(\sqrt[4]{16x}\):

    • The fourth root involves variables with undergo an absolute value due to even roots: \[ \sqrt[4]{16x} = \sqrt[4]{16} \cdot \sqrt[4]{x} = 2\sqrt[4]{x}. \]
    • Here, since we are dealing with an even root, it requires an absolute value for \(x\): \[ = 2\sqrt[4]{|x|} \text{ (not simplified as it should be)}. \]
    • However, in this particular notation, the absolute value isn’t explicitly written.
  4. \(\sqrt[3]{-8x^3}\):

    • The third root of \( -8x^3 \) simplifies directly to: \[ \sqrt[3]{-8x^3} = \sqrt[3]{-8} \cdot \sqrt[3]{x^3} = -2x. \]
    • No absolute value is introduced here because \(x^3\) retains its sign.

Conclusion: Among the options given, \(\sqrt[4]{16x}\) is the only expression that would introduce the need for an absolute value to ensure the expression remains valid for all \(x\), but it's not written showing the absolute value properly. Therefore, if we refer explicitly to needing absolute values:

  • \(\sqrt[4]{16x}\) has an implicit need for absolute value but is not written with it.

In strict terms of this question, it is often the case with even roots generating expressions requiring absolute values as seen.