Question

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

• A. \(4^{3x^{2}}..... 5^{y}..... 3^{z^{4}}\)
• B. \(4^{3x^{2}}..... 5^{y/2}..... 9z^{2}\)
• C. \(4^{6x}..... 25^{y/4}..... 9z^{2}\)
• D. \(4^{3x^{2}}..... 5^{y/2}..... 3\,z^{2}\)

Hint:

\(\sqrt{a^{m}}=a^{m/2}\). For bases like \(25\) and \(9\), rewrite as powers of primes (\(5^2, 3^2\)) before halving the exponent.

Answer 1

The Correct Option is D

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