If $x = \frac{\sqrt{3} + 1}{\sqrt{2} + 1}$, what is $\frac{\sqrt{3} - 1}{\sqrt{2} - 1}$ in terms of $x$?

Question

I could not manage to solve this easy question on my test book. Sorry the image is in Turkish.

I will try to translate it, it says if $$x = \frac{\sqrt{3} + 1}{\sqrt{2} + 1}$$ what is $$\frac{\sqrt{3} - 1}{\sqrt{2} - 1}$$ in terms of $x$?

I do not know if I translated it well, but thanks for any help you can provide anyway.

Answer 1

$$x=\frac{\sqrt 3+1}{\sqrt 2+1}$$ Multiply both sides by $\frac{\sqrt 3-1}{\sqrt 2-1}$ $$x\cdot\frac{\sqrt 3-1}{\sqrt 2-1}=\frac{\sqrt 3+1}{\sqrt 2+1}\cdot\frac{\sqrt 3-1}{\sqrt 2-1}$$ Use that $(a+b)(a-b)=a^2-b^2$ $$x\cdot\frac{\sqrt 3-1}{\sqrt 2-1}=\frac{3-1}{2-1}=2$$ Finally, divide by $x$ $$\frac{\sqrt 3-1}{\sqrt 2-1}=\frac{2}{x}$$

Answer 2

Hint: Multiply $$x = \frac{\sqrt{3} + 1}{\sqrt{2} + 1}$$ by $$\frac{\sqrt{3} - 1}{\sqrt{3} - 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1}$$ then compare the result with $$\frac{\sqrt{3} - 1}{\sqrt{2} - 1}$$

Answer 3

Let ${\displaystyle x:={\sqrt{3}+1\over\sqrt{2}+1}}$ and ${\displaystyle y:={\sqrt{3}-1\over\sqrt{2}-1}}$. Then ${\displaystyle x\cdot y={2\over1}}$, hence ${\displaystyle y={2\over x}}$.