How do you rationalize the denominator and simplify #sqrt(3x)/(sqrtx-sqrt3)#?

1 Answer
smendyka
Jul 5, 2017

See a solution process below:

Explanation:

To rationalize the denominator multiply the expression by this form of #1#:

#(sqrt(x) + sqrt(3))/(sqrt(x) + sqrt(3))#

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

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

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

#(sqrt(3x)(sqrt(x) + sqrt(3)))/(x - 0 - 3) =>#

#(sqrt(3x)(sqrt(x) + sqrt(3)))/(x - 3) =>#

Now, we can work on simplifying the numerator:

#(color(red)(sqrt(3x))(sqrt(x) + sqrt(3)))/(x - 3) =>#

#((color(red)(sqrt(3x)) xx sqrt(x)) + (color(red)(sqrt(3x)) xx sqrt(3)))/(x - 3) =>#

#(sqrt(3x^2) + sqrt(9x))/(x - 3) =>#

#(sqrt(3)x + 3sqrt(x))/(x - 3)#

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