How do you rationalize the denominator and simplify #sqrtx/(sqrtx+sqrt3)#?

1 Answer
Jun 30, 2016

Answer:

#sqrt(x)/(sqrt(x)+sqrt(3))=color(blue)((x-sqrt(3x))/(x-3))#

Explanation:

For the general expression: #(a+b)#
the conjugate is #(a-b)#
and the product of an expression times its conjugate is
#color(white)("XXX")(a+b)*(a-b)=a^2-b^2#

If #a# and/or #b# are square roots this lets us get rid of the square roots.

For the given example #sqrt(x)/(sqrt(x)+sqrt(3))#

multiplying both the numerator and denominator by the conjugate of #(sqrt(x)+sqrt(3))# lets us remove the square roots from the denominator.

#color(white)("XXX")sqrt(x)/(sqrt(x)+sqrt(3))xx(sqrt(x)-sqrt(3))/(sqrt(x)-sqrt(3))#

#color(white)("XXX")=(sqrt(x)*(sqrt(x)-sqrt(3)))/(x-3)#

#color(white)("XXX")=(x-sqrt(3x))/(x-3)#