1. ## surds

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

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

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

Thanks .

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

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

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

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

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

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

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

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

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

Thanks .
Hi

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

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