Question 449279
{{{-12/sqrt(18x)}}}
<pre>
Don't start out multiplying by the entire denominator over itself.
Instead, first get the prime factorization of 18 as 2*3²

{{{-12/sqrt(2*3^2*x)}}}

To get a perfect square under the radical,
we need to multiply what's under there by another 2 and also
by another x, so we multiply by {{{sqrt(2x)/sqrt(2x)}}} which is just 1 in value.

{{{(-12/sqrt(2*3^2*x))(sqrt(2x)/sqrt(2x))}}}

Multiplying under the radicals in the denominator:

{{{(-12sqrt(2x))/sqrt(2^2*3^2*x^2)}}}

Now we will no longer have a radical in the
denominator because we can take the square roots
of all those squares

{{{(-12sqrt(2x))/(2*3*x)}}}

{{{(-12sqrt(2x))/(6x)}}}

Then the 6 divides into the 12 and we have:

{{{(-2sqrt(2x))/x}}}

Edwin</pre>