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6. critique reasoning kara says that $(-4)^{\frac{1}{2}} = sqrt{(-4)^1}…

Kara is not correct. 1. $(-4)^{\frac{1}{2}}$ is not a real number (it is imaginary, $2i$), so it cannot equal a real number like 2. 2. $\sqrt{(-4)^2}=\sqrt{16}=4$, not 2. 3. The r…

カテゴリ: algebra 更新日: 2026-02-09

問題

  1. critique reasoning kara says that $(-4)^{\frac{1}{2}} = sqrt{(-4)^1} = sqrt{16} = 2$. is she correct? explain why or why not.

解法ステップ

  1. Understand the question
    1. critique reasoning kara says that $(-4)^{\frac{1}{2}} = sqrt{(-4)^1} = sqrt{16} = 2$. is she correct? explain why or why not.
  2. Explanation

    Step1: Evaluate left-hand side

    $(-4)^{\frac{1}{2}}$ represents the square root of $-4$. In real numbers, $\sqrt{-4}$ is not a real number (it is the imaginary number $2i$).

    Step2: Evaluate middle expression

    $\sqrt{(-4)^2} = \sqrt{16}$. The square root of a non-negative number gives the non-negative root, so $\sqrt{16}=4$.

    Step3: Evaluate right-hand side

    $\sqrt{16}=4$, not 2. Also, the initial step of equating $(-4)^{\frac{1}{2}}$ to $\sqrt{(-4)^2}$ is invalid because $a^{\frac{1}{2}}$ is defined as $\sqrt{a}$ only when $a\geq0$ for real numbers.

  3. Final answer

    Kara is not correct.

    1. $(-4)^{\frac{1}{2}}$ is not a real number (it is imaginary, $2i$), so it cannot equal a real number like 2.
    2. $\sqrt{(-4)^2}=\sqrt{16}=4$, not 2.
    3. The rule $a^{\frac{1}{2}}=\sqrt{a^2}$ does not hold for negative $a$ in the real number system, as square roots of negative numbers are not real, and $\sqrt{a^2}=|a|$ for real $a$.

答え

Explanation

Step1: Evaluate left-hand side

$(-4)^{\frac{1}{2}}$ represents the square root of $-4$. In real numbers, $\sqrt{-4}$ is not a real number (it is the imaginary number $2i$).

Step2: Evaluate middle expression

$\sqrt{(-4)^2} = \sqrt{16}$. The square root of a non-negative number gives the non-negative root, so $\sqrt{16}=4$.

Step3: Evaluate right-hand side

$\sqrt{16}=4$, not 2. Also, the initial step of equating $(-4)^{\frac{1}{2}}$ to $\sqrt{(-4)^2}$ is invalid because $a^{\frac{1}{2}}$ is defined as $\sqrt{a}$ only when $a\geq0$ for real numbers.

Answer

Kara is not correct.

  1. $(-4)^{\frac{1}{2}}$ is not a real number (it is imaginary, $2i$), so it cannot equal a real number like 2.
  2. $\sqrt{(-4)^2}=\sqrt{16}=4$, not 2.
  3. The rule $a^{\frac{1}{2}}=\sqrt{a^2}$ does not hold for negative $a$ in the real number system, as square roots of negative numbers are not real, and $\sqrt{a^2}=|a|$ for real $a$.

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Question Analysis

Subject unspecified
Sub Subject unspecified
Education Level high school
Difficulty unspecified
Question Type text only
Multi Question No
Question Count 1
Analysis Status completed
Analyzed At 2026-02-09T19:56:58

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6. critique reasoning  kara says that $(-4)^{\frac{1}{2}} = sqrt{(-4)^1} = sqrt{16} = 2$. is she correct? explain why or why not.

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