\(x=735\)

\(735a=perfect\) \(square\)

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!