# Surds problem

• Mar 5th 2014, 08:39 AM
grillage
Surds problem
This is a question from a local past paper. I am either having a mental breakdown or there is a problem in the question. As usual I need a sanity check please. I don't necessarily need the answer, but if you can tell me that it is solveable and there is not a mistake, then that is good.

Using surds, find the values of a and b if $\frac{\sqrt{3}+a}{\sqrt{3}+b}=\sqrt{3}-1$
• Mar 5th 2014, 09:01 AM
Plato
Re: Surds problem
Quote:

Originally Posted by grillage
This is a question from a local past paper. I am either having a mental breakdown or there is a problem in the question. As usual I need a sanity check please. I don't necessarily need the answer, but if you can tell me that it is solveable and there is not a mistake, then that is good.

Using surds, find the values of a and b if $\frac{\sqrt{3}+a}{\sqrt{3}+b}=\sqrt{3}-1$

$a=-1 ~\&~ b=1-\sqrt{3}$ works.
• Mar 5th 2014, 09:03 AM
Soroban
Re: Surds problem
Hello, grillage!

Quote:

$\text{Find the values of }a\text{ and }b\text{ if: }\:\frac{\sqrt{3}+a}{\sqrt{3}+b}\:=\:\sqrt{3}-1$

We have: . $\sqrt{3} + a \;=\;(\sqrt{3}-1)(\sqrt{3}+b)$

. . . . . . . . $\sqrt{3}+a \;=\;3 + b\sqrt{3} - \sqrt{3} - b$

n . . . $a + b - b\sqrt{3} \;=\;3 - 2\sqrt{3}$

Equate rational and irreational components: . $\begin{Bmatrix}a + b \:=\:3 \\ -b\sqrt{3} \:=\:-2\sqrt{3}\end{Bmatrix}$
Solve the system: . $\begin{Bmatrix}a \:=\:1 \\ b \:=\:2\end{Bmatrix}$
• Mar 5th 2014, 09:06 AM
ebaines
Re: Surds problem
Multiply through by the denominator to get:

$\sqrt 3 +a = 3 - \sqrt 3 +b(\sqrt 3 -1)$

and rearrange to get b in terms of a:

$b = \frac {a -3 + 2 \sqrt 3 }{\sqrt 3 - 1}$

Now to pretty it up you can multiply through by $( \sqrt 3 + 1)/(\sqrt 3 + 1)$ to get rid of the square root in the denominator. This gives an equation for 'b' as a function of 'a'. You should find for example that if 'a' = 1 then 'b' = 2, and if 'a' = -1 then b = 1-sqrt(3).