Rationalise the denominator with two surds? (1 Viewer)

Rohanco · Oct 2, 2012

Having trouble here;

Express as a single fraction with rational denominator...

Thanks

Demise · Oct 2, 2012

Curious as to why you multiplied by 3?

HSC2014 · Oct 2, 2012

You can rationalise the bottom by multiplying both the top and bottom by 2*root10 - 1
I don't know the proper term of it but you will be creating a difference of two squares (a^2 - b^2) in the denominator and thus removing the surd.

gr_111 · Oct 2, 2012

HSC2014 said:
You can rationalise the bottom by multiplying both the top and bottom by 2*root10 - 1
I don't know the proper term of it but you will be creating a difference of two squares (a^2 - b^2) in the denominator and thus removing the surd.

+1. You got as far as simplifying it, now you just have to rationalise it.

Drongoski · Oct 2, 2012

$= \frac {1}{\sqrt 5 + \sqrt 2} \times \frac {\sqrt 5 - \sqrt 2} {\sqrt 5 - \sqrt 2} + \frac {3}{3\sqrt2 - \sqrt 5} \times \frac {3\sqrt 2 + \sqrt 5}{3\sqrt 2 + \sqrt 5}$

$= \frac {\sqrt 5 - \sqrt 2}{(\sqrt 5)^2 - (\sqrt 2)^2} + \frac {3(3\sqrt 2 + \sqrt 5)}{(3\sqrt 2)^2 - (\sqrt 5)^2}$

$= \frac {\sqrt 5 - \sqrt 2}{3} + \frac {3(3\sqrt 2 + \sqrt 5)}{13}$

.

.

$= \frac {22\sqrt 5 + 14\sqrt 2}{39}$

Rohanco · Oct 2, 2012

Drongoski said:
$= \frac {22\sqrt 5 + 14\sqrt 2}{39}$

Thanks so much

Rationalise the denominator with two surds? (1 Viewer)

Rohanco

Member

Attachments

Demise

Member

HSC2014

Member

gr_111

Member

Drongoski

Well-Known Member

Rohanco

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)