How do you simplify #\frac { \sqrt { a } } { \sqrt { a } - \sqrt { n } }#?

2 Answers
George C.
Oct 25, 2017

A couple of ideas...

Explanation:

Given:

#sqrt(a)/(sqrt(a)-sqrt(n))#

I am not sure what we would consider a simplified expression, but here are some things we can do:

Option 1a - Divide numerator and denominator by #sqrt(a)#

#sqrt(a)/(sqrt(a)-sqrt(n)) = 1/(1-sqrt(n)/sqrt(a)) = 1/(1-sqrt(n/a))#

Option 1b - Multiply both numerator and denominator by #sqrt(a)#

#sqrt(a)/(sqrt(a)-sqrt(n)) = a/(a-sqrt(a)sqrt(n)) = a/(a-sqrt(an))#

Option 2 - Rationalise the denominator

We can rationalise the denominator by multiplying both numerator and denominator by the radical conjugate of the denominator...

#sqrt(a)/(sqrt(a)-sqrt(n)) = (sqrt(a)(sqrt(a)+sqrt(n)))/((sqrt(a)-sqrt(n))(sqrt(a)+sqrt(n)))#

#color(white)(sqrt(a)/(sqrt(a)-sqrt(n))) = (a+sqrt(a)sqrt(n))/(a-n)#

#color(white)(sqrt(a)/(sqrt(a)-sqrt(n))) = (a+sqrt(an))/(a-n)#

Jumbotron
Oct 25, 2017

#(a + sqrt(an))/(a - n)#

Explanation:

That's conjugate surd..

See Process Below;

#sqrta/(sqrta - sqrtn)#

#sqrta/(sqrta - sqrtn) xx color(blue)((sqrta + sqrtn)/(sqrta + sqrtn)) -> "Conjugate"#

#(sqrta (sqrta + sqrtn))/((sqrta - sqrtn) (sqrta + sqrtn))#

#(a+ (sqrt(a xx n)))/(a - n)#

#(a + sqrt(an))/(a - n) -> "Simplified"#

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