# [SOLVED] Find the values of p and q

• Apr 27th 2008, 01:43 AM
anne.
[SOLVED] Find the values of p and q
(2√3 - 1)² = p + q√3

Does anyone know how to solve this?
So far I've only got:

13 - 4√3 = p + q√3

However, as with a similar problem I have done, I am pretty sure there are a range of answers, not just p = 13 and q = -4.

Can someone help me with this?
• Apr 27th 2008, 01:49 AM
Moo
Hello,

It may depend on how you define p and q. Do they belong to $\displaystyle \mathbb{N}$, or $\displaystyle \mathbb{R}$ ?
• Apr 27th 2008, 01:57 AM
Isomorphism
Quote:

Originally Posted by Moo
Hello,

It may depend on how you define p and q. Do they belong to $\displaystyle \mathbb{N}$, or $\displaystyle \mathbb{R}$ ?

Or $\displaystyle \mathbb{Q}$, ya know, the rationals :)
• Apr 27th 2008, 01:58 AM
Moo
If it belongs to $\displaystyle \mathbb{Q}$ there can't be several solutions, just one :)

By the way, I meant $\displaystyle \mathbb{Z}$, certainly not $\displaystyle \mathbb{N}$
• Apr 27th 2008, 01:59 AM
anne.
The question did not contain any further detail, also, I do not know what $\displaystyle \mathbb{N}$ or $\displaystyle \mathbb{R}$ are...we haven't learned it, so I don't think they belong to....those things. (Itwasntme)

Erm....could someone explain what those large, nice looking letters mean? (Worried)

Thanks. (Bow)(Bow)(Bow)
• Apr 27th 2008, 02:03 AM
Moo
$\displaystyle \mathbb{N}$ : set of natural integers (means the positive ones)

$\displaystyle \mathbb{Z}$ : set of all integers (positive and negative)

$\displaystyle \mathbb{Q}$ : rational numbers, which means $\displaystyle \frac pq$ with p and q in $\displaystyle \mathbb{Z}$ (and q different from 0)

$\displaystyle \mathbb{R}$ : set of all numbers, including irrational ones.

$\displaystyle \mathbb{N}$ is included in $\displaystyle \mathbb{Z}$, which is included in $\displaystyle \mathbb{Q}$, which is included in $\displaystyle \mathbb{R}$

-----------------------------

Why did I ask ? Because if we're working on $\displaystyle \mathbb{Z}$, there is only one solution.

If we're working on $\displaystyle \mathbb{R}$, then $\displaystyle p=10-5\sqrt{3}$ and $\displaystyle q=1+\sqrt{3}$ are solutions, for example, but there is an infinity of possibilities. (note that $\displaystyle \sqrt{3}$ is an irrational and only belongs to $\displaystyle \mathbb{R}$).
• Apr 27th 2008, 02:08 AM
Isomorphism
Quote:

Originally Posted by Moo
If it belongs to $\displaystyle \mathbb{Q}$ there can't be several solutions, just one :)

By the way, I meant $\displaystyle \mathbb{Z}$, certainly not $\displaystyle \mathbb{N}$

In the most general sense, shouldnt you be saying $\displaystyle \mathbb{Q}$, since they include $\displaystyle \mathbb{Z}$(this is where uniqueness of the representation holds)?
• Apr 27th 2008, 02:09 AM
anne.
Well the question did not state anything, so I'd assume the solution would belong to $\displaystyle \mathbb{R}$.
• Apr 27th 2008, 02:17 AM
Moo
So the solutions will be

$\displaystyle p=a+b\sqrt{3}$ and $\displaystyle q=c+d\sqrt{3}$

$\displaystyle p+q\sqrt{3}=a+b\sqrt{3}+c\sqrt{3}+3d$

So we want $\displaystyle a+3d=13$ and $\displaystyle b+c=-4$

Hence a=13-3d and b=-4-c.

-------> general solutions are $\displaystyle p=13-3d-(4+c)\sqrt{3}$ and $\displaystyle q=c+d \sqrt{3}$, this time, c and d belonging to $\displaystyle \mathbb{Q}$ (yep Isomorphism, it can be done this way indeed :))

This is only for the ones involving $\displaystyle \sqrt{3}$ the basic way. Otherwise, there is an infinity of them.
• Apr 27th 2008, 02:19 AM
Moo
Since they ask you the "values of p and q", I'd follow the idea that the solutions are simply p=13 and q=-4
• Apr 27th 2008, 02:20 AM
Isomorphism
Quote:

Originally Posted by anne.
Well the question did not state anything, so I'd assume the solution would belong to $\displaystyle \mathbb{R}$.

Then you have $\displaystyle \mathbb{R}$ solutions :)

Pick any real number. Let it be p. Substitute in your equation and solve for q, another real number!

From my experience, I think you are asked to find rational values(that is the numbers from $\displaystyle \mathbb{Q}$ and not $\displaystyle \mathbb{R}$).

Doing it for $\displaystyle \mathbb{R}$ looks uneducational (Thinking)
• Apr 27th 2008, 02:30 AM
anne.
O_O;;

wow. I didn't get a word of the second-last post that you posted (the one with all the equations...)

I'll be sticking with the simple solution of p = 13 and q = -4 then.