# [SOLVED] Rationalising Denominators.

• Feb 1st 2009, 06:59 AM
Halo
[SOLVED] Rationalising Denominators.
Okay, so this is the question...

(2+√3) / (1+√2)

This is what I have so far.

(2+√3) (1-√2) / (1+√2) (1-√2)

= 2 - 2√2 + √3 - √3√2
1 - √2 + √2 - √2√2

Not sure where to go from here?

Thanks.

Edit: Sorry, the question on my worksheet is, "Rationalise the denominators of the following..."
• Feb 1st 2009, 07:20 AM
Re :
Quote:

Originally Posted by Halo
Okay, so this is the question...

(2+√3) / (1+√2)

This is what I have so far.

(2+√3) (1-√2) / (1+√2) (1-√2)

= 2 - 2√2 + √3 - √3√2
1 - √2 + √2 - √2√2

Not sure where to go from here?

Thanks.

Edit: Sorry, the question on my worksheet is, "Rationalise the denominators of the following..."

$\displaystyle \frac{2+\sqrt{3}}{1+\sqrt{2}}\times\frac{1-\sqrt{2}}{1-\sqrt{2}}$

$\displaystyle -(2+\sqrt{3})(1-\sqrt{2})$
• Feb 1st 2009, 07:29 AM
princess_21
Quote:

Originally Posted by Halo
Okay, so this is the question...

(2+√3) / (1+√2)

This is what I have so far.

(2+√3) (1-√2) / (1+√2) (1-√2)

= 2 - 2√2 + √3 - √3√2
1 - √2 + √2 - √2√2

after that step try to simplify the problem
2 - 2√2 + √3 - √3√2
1 - √2 + √2 - √2√2

2 -2√2 + √3 - √6
-1

or

$\displaystyle -(2-2\sqrt{2}+\sqrt{3}-\sqrt{6})$
• Feb 2nd 2009, 02:21 AM
Re :
Thanks Jhevon , my mistake .
• Feb 2nd 2009, 09:41 AM
Bruce
Where did you get the minus sign from?

• Feb 3rd 2009, 03:26 AM
princess_21
Quote:

Originally Posted by Bruce
Where did you get the minus sign from?

$\displaystyle \frac{2+\sqrt{3}}{1+\sqrt{2}}$

then i rationalized it by multiplying $\displaystyle (1-\sqrt{2})$

$\displaystyle \frac{2+\sqrt{3}}{1+\sqrt{2}} X \frac{1+\sqrt{2}}{1+\sqrt{2}}$

$\displaystyle \frac{2-2\sqrt{2}+\sqrt{3}-\sqrt{6}}{1-\sqrt{2}+\sqrt{2}-\sqrt{2}\sqrt{2}}$

$\displaystyle \frac{2-\sqrt{2}+\sqrt{3}+\sqrt{6}}{1-4}$

$\displaystyle \frac{2-\sqrt{2}+\sqrt{3}+\sqrt{6}}{-1}$

$\displaystyle -(2-\sqrt{2}+\sqrt{3}+\sqrt{6})]$

$\displaystyle -(2+\sqrt{3})(1-\sqrt{2})$

i have the same answer as Jhevon i just simplified it by multiplying.