How do you rationalize the denominator and simplify #6/(sqrt(2x))#?

2 Answers
Apr 1, 2018

Answer:

#[3sqrt(2x)]/x#

Explanation:

#6/(sqrt(2x))#
#6/(sqrt(2x))*[(cancel-sqrt(2x))/(cancel(-)sqrt(2x))]#
which will give,
#(cancel(6)sqrt(2x))/(cancel(2)x)=[3sqrt(2x)]/x#

Apr 1, 2018

Answer:

#(3sqrt(2x))/x#

Explanation:

#"note that "sqrtaxxsqrta=a#

#"to rationalise the denominator, that is eliminate the"#
#"radical from the denominator then"#

#"multiply numerator/denominator by "sqrt(2x)#

#rArr6/(sqrt(2x))xxsqrt(2x)/sqrt(2x)#

#=(6sqrt(2x))/(2x)=(3sqrt(2x))/x#