# Math Help - surds...

1. ## surds...

hiya all
theres this question which i can't seem to get the right answer to...
3+ square root of 24 / 2 + square root of 6
you need to put it in the form of a + b square root of c / d

the answer is 6- square root of 6 / 2 buut i can't get it!!

thaank you

2. Originally Posted by katiethegreat
hiya all
theres this question which i can't seem to get the right answer to...
3+ square root of 24 / 2 + square root of 6
you need to put it in the form of a + b square root of c / d

the answer is 6- square root of 6 / 2 buut i can't get it!!

thaank you
Multiply by it's conjugate which is $2-\sqrt6$

$\frac{3+2\sqrt6}{2+\sqrt6} \times \frac{2-\sqrt6}{2-\sqrt6} = \frac{(3+2\sqrt6)(2-\sqrt6)}{(2+\sqrt6)(2-\sqrt6)} = \frac{6-3\sqrt6+4\sqrt6 -12}{4-6}$

I will leave you to simplify that

3. thank you

4. Originally Posted by katiethegreat
hiya all
theres this question which i can't seem to get the right answer to...
3+ square root of 24 / 2 + square root of 6
you need to put it in the form of a + b square root of c / d

the answer is 6- square root of 6 / 2 buut i can't get it!!

thaank you
Hi katiethegreat,

$\frac{3+\sqrt{24}}{2+\sqrt{6}}=\frac{3+\sqrt{6(4)} }{2+\sqrt{6}}=\frac{3+\sqrt{6}\sqrt{4}}{2+\sqrt{6} }=\frac{3+2\sqrt{6}}{2+\sqrt{6}}$

To get 2 under the line instead of a surd, multiply the entire fraction by 1,
in such a way, that you get an integer under the line.
This means getting rid of the surd part of the denominator.

To do this, change the sign of one of the 2 values in the denominator to form this fraction.

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

Alternatively,

$\frac{3+2\sqrt{6}}{2+\sqrt{6}}=p+q\sqrt{6}$

$3+2\sqrt{6}=(2+\sqrt{6})(p+q\sqrt{6})=2(p+q\sqrt{6 })+\sqrt{6}(p+q\sqrt{6})$

$=2p+2q\sqrt{6}+p\sqrt{6}+q(6)=(2p+6q)+(p+2q)\sqrt{ 6}$

$2p+6q=3$

$p+2q=2\ \Rightarrow\ 2p+4q=4$

Solving the simultaneous equations gves

$2q=-1\ \Rightarrow\ q=-\frac{1}{2}$

$p=3$

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