If $x=735$ and $ax$ is a perfect square where $a$ is a positive integer, what is the smallest possible value of $\sqrt{ax}$?
+363 \(x=735\)
\(735a=perfect\) \(square\)
well, \(735=7\times105=7\times5\times3\times7\)
\(735=3\times5\times7^2\)
in order to get a perfect square, you will need a 3 and a 5 because every number in the prime factorization needs to have a square.
\(\sqrt{735\cdot15}=\sqrt{3^2\cdot5^2\cdot7^2}=3\cdot5\cdot7=105\)
notice that you can take a shortcut by multiplying 3x5x7.
I'll let you wonder why, but it works all the time!
+363 \(x=735\)
\(735a=perfect\) \(square\)
well, \(735=7\times105=7\times5\times3\times7\)
\(735=3\times5\times7^2\)
in order to get a perfect square, you will need a 3 and a 5 because every number in the prime factorization needs to have a square.
\(\sqrt{735\cdot15}=\sqrt{3^2\cdot5^2\cdot7^2}=3\cdot5\cdot7=105\)
notice that you can take a shortcut by multiplying 3x5x7.
I'll let you wonder why, but it works all the time!
