# Rationalise the denominator and simplify

• December 26th 2008, 09:29 PM
Joker37
Rationalise the denominator and simplify
1/2√7 + 2√3 - 1/3√7 + √3
• December 26th 2008, 09:58 PM
nzmathman
Is that $\frac{1}{2\sqrt{7}} + 2\sqrt{3} - \frac{1}{3\sqrt{7}} + \sqrt{3}$

or $\frac{1}{2} \sqrt{7} + 2\sqrt{3} - \frac{1}{3} \sqrt{7} + \sqrt{3}$ ?

I'll assume its the first one since we are supposed to rationalise a denominator.

$\frac{1}{2\sqrt{7}} + 2\sqrt{3} - \frac{1}{3\sqrt{7}} + \sqrt{3}$

$=\frac{3 - 2 +3\sqrt{3} \times 6\sqrt{7}}{6\sqrt{7}}$

$=\frac{1 +18\sqrt{21} }{6\sqrt{7}}$

Now times top and bottome by $\sqrt{7}$ to rationalise the denominator.
• December 26th 2008, 09:58 PM
sudha_14
easy.....
I think it's :
$\frac {1} {2\sqrt{7} + 2\sqrt {3}} - \frac {1} {3\sqrt {7} + \sqrt {3}}$

right??
• December 26th 2008, 10:03 PM
Joker37
Sorry, I meant:

1/(2√7 + 2√3) - 1/(3√7 + √3)

LOL, I was thinking of putting brackets but hesitated.
• December 26th 2008, 10:08 PM
Joker37
Quote:

Originally Posted by sudha_14
I think it's :
$\frac {1} {2\sqrt{7} + 2\sqrt {3}} - \frac {1} {3\sqrt {7} + \sqrt {3}}$

right??

Yep!
• December 26th 2008, 10:15 PM
nzmathman
$\frac {1} {2\sqrt{7} + 2\sqrt {3}} - \frac {1} {3\sqrt {7} + \sqrt {3}}$

$= \frac{3\sqrt {7} + \sqrt {3} - 2\sqrt{7} - 2\sqrt{3}}{(2\sqrt{7} + 2\sqrt {3})(3\sqrt {7} + \sqrt {3})}$

$= \frac{\sqrt {7} - \sqrt {3}}{(2\sqrt{7} + 2\sqrt {3})(3\sqrt {7} + \sqrt {3})}$

Now to rationalise the denominator times the fraction (bottom and top) by the conjugates of the two surds to make the two factors rational.

$= \frac{(\sqrt {7} - \sqrt {3})(2\sqrt{7} - 2\sqrt {3})(3\sqrt {7} - \sqrt {3})}{(2\sqrt{7} + 2\sqrt {3})(2\sqrt {7} - 2\sqrt {3})(3\sqrt{7} + \sqrt {3})(3\sqrt {7} - \sqrt {3})}$

$= \frac{(\sqrt {7} - \sqrt {3})(2\sqrt{7} - 2\sqrt {3})(3\sqrt {7} - \sqrt {3})}{(28 - 12)(63 - 3)}$

$= \frac{(\sqrt {7} - \sqrt {3})(2\sqrt{7} - 2\sqrt {3})(3\sqrt {7} - \sqrt {3})}{16 \times 60}$

$= \frac{(\sqrt {7} - \sqrt {3})(2\sqrt{7} - 2\sqrt {3})(3\sqrt {7} - \sqrt {3})}{960}$

Now, I leave it up to you to expand the top...
• December 26th 2008, 10:32 PM
sudha_14