Given $(1-2k)x^2 - (3k+4)x + 2 = 0$ for some $k \in \mathbb{R}\setminus\{1/2\}$, suppose $x\_1$ and $x\_2$ are distinct roots of the equation such that $x\_1 x\_2 = 1$.

Without using Vieta's formulas, how can we show $k = -1/2$ ?

Here is what I have done so far:

$(1-2k)x\_1^2 - (3k+4)x\_1 + 2 = 0$

$(1-2k)x\_2^2 - (3k+4)x\_2 + 2 = 0$

$\to (1-2k)x\_1^2 - (3k+4)x\_1 + 2 = (1-2k)x\_2^2 - (3k+4)x\_2 + 2$

$\to (1-2k)x\_1^2 - (3k+4)x\_1 = (1-2k)x\_2^2 - (3k+4)x\_2$

$\to (1-2k)x\_1^2 - (3k+4)x\_1 = (1-2k)(1/x\_1)^2 - (3k+4)(1/x\_1)$

$\to (1-2k)[x\_1^2 - (1/x\_1)^2] - (3k+4)[x\_1 - (1/x\_1)] = 0$

$\to (1-2k)[x\_1 - (1/x\_1)][x\_1 + (1/x\_1)] - (3k+4)[x\_1 - (1/x\_1)] = 0$

$\to (1-2k)[x\_1 + (1/x\_1)] - (3k+4) = 0$ or $[x\_1 - (1/x\_1)] = 0$

$\to (1-2k)[x\_1 + (1/x\_1)] - (3k+4) = 0$, $x\_1 = 1$ or $x\_1 = -1$

Since the later two cases violate the distinct roots assumption, we have:

$(1-2k)[x\_1 + (1/x\_1)] - (3k+4) = 0$

This gives:

[![enter image description here](https://i.stack.imgur.com/izSKj.png)](https://i.stack.imgur.com/izSKj.png)

The [answer](http://www.wolframalpha.com/input/?i=%281-2k%29x%5E2%20-%283k%2B4%29x%20%2B%202%20%3D%200%2C%20x%3D1%2Fy%2C%20%281-2k%29y%5E2%20-%283k%2B4%29y%20%2B%202%20%3D%200) is $x\_1 = \frac{5-i\sqrt{39}}{8}$ or $x\_1 = \frac{5+i\sqrt{39}}{8}$. How do I get that (without Vieta's)?