Question:

Find \(\sqrt{4^{6x^{2}} 25^{y/2} 9 z^{4}}\).

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\(\sqrt{a^{m}}=a^{m/2}\). For bases like \(25\) and \(9\), rewrite as powers of primes (\(5^2, 3^2\)) before halving the exponent.
Updated On: Aug 25, 2025
  • \(4^{3x^{2}} 5^{y} 3^{z^{4}}\)
  • \(4^{3x^{2}} 5^{y/2} 9z^{2}\)
  • \(4^{6x} 25^{y/4} 9z^{2}\)
  • \(4^{3x^{2}} 5^{y/2} 3z^{2}\)
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The Correct Option is D

Solution and Explanation

\(\sqrt{4^{6x^{2}}}=4^{3x^{2}}\), \(\sqrt{25^{y/2}}=25^{y/4}=(5^{2})^{y/4}=5^{y/2}\),
\(\sqrt{9}=3\), \(\sqrt{z^{4}}=z^{2}\) (assuming \(z\ge 0\)).
Multiply: \(4^{3x^{2}} 5^{y/2} 3 z^{2}\).
\[ \boxed{4^{3x^{2}} 5^{y/2} 3z^{2}} \]
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