# Math Help - surds

1. ## surds

Express $\sqrt{66-24\sqrt{6}}$in the form of $p\sqrt{2}+q\sqrt{3}$

I got $p=-3$ and $q=4$ or $p=3$and $q=-4$

The answer is $p=-3$ and $q=4$ . My question is why is $p=3$and $q=-4$ ommitted ??

Thanks .

2. Hello,
Originally Posted by thereddevils
Express $\sqrt{66-24\sqrt{6}}$in the form of $p\sqrt{2}+q\sqrt{3}$

I got $p=-3$ and $q=4$ or $p=3$and $q=-4$

The answer is $p=-3$ and $q=4$ . My question is why is $p=3$and $q=-4$ ommitted ??

Thanks .
That's because $\sqrt{66-24\sqrt{6}}$ is positive (it's a square root.. if it was $-\sqrt{66-24\sqrt{6}}$, it would surely be negative)

Now, have a thought on this...
$\sqrt{2}<\sqrt{3}$
Then
$3\sqrt{2}<3\sqrt{3}<4\sqrt{3}$

So if p=3 and q=-4, would $p\sqrt{2}+q\sqrt{3}$ be positive ??

3. Originally Posted by thereddevils
Express $\sqrt{66-24\sqrt{6}}$in the form of $p\sqrt{2}+q\sqrt{3}$

I got $p=-3$ and $q=4$ or $p=3$and $q=-4$

The answer is $p=-3$ and $q=4$ . My question is why is $p=3$and $q=-4$ ommitted ??

Thanks .
Hi

You effectively get 2 possible solutions since $\left(3\sqrt{2}-4\sqrt{3}\right)^2 = \left(-3\sqrt{2}+4\sqrt{3}\right)^2 = 66-24\sqrt{6}$

But $\sqrt{66-24\sqrt{6}} > 0$ and $-3\sqrt{2}+4\sqrt{3} > 0$ but $3\sqrt{2}-4\sqrt{3} < 0$