If $x=735$ and $ax$ is a perfect square where $a$ is a positive integer, what is the smallest possible value of $\sqrt{ax}$?

Guest Mar 1, 2022

#1**+4 **

\(x=735\)

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

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!

XxmathguyxX Mar 1, 2022

#1**+4 **

Best Answer

\(x=735\)

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

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!

XxmathguyxX Mar 1, 2022