Related to this answer: Find k in $(1−2k)x^2−(3k+4)x+2=0$ given facts about the roots.
In the partial fraction decomposition of $\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$, we have:
$0x + 1 = A(x+1) + B(x-1) \to A+B = 0$, $A - B = 1$, where we have matched coefficients.
On the other hand, suppose I was given a quadratic equation
$$(1−2k)x^2−(3k+4)x+2=0$$
and said that its two roots are
$$x_1 = 1, x_2 = 1$$.
It is clear that $k = -1/5$.
However, if $x_1 = 1, x_2 = 1$, it is correct to say that $x^2 - 2x + 1 = 0$.
Since $(1−2k)x^2−(3k+4)x+2=0$ is quadratic, $(1−2k)\ne0$.
$\to x^2−\frac{(3k+4)}{(1−2k)}x+\frac{2}{(1−2k)}=0$
$\to x^2 - 2x + 1 = 0 = x^2−\frac{(3k+4)}{(1−2k)}x+\frac{2}{(1−2k)}$
$\to x^2 - 2x + 1 = x^2−\frac{(3k+4)}{(1−2k)}x+\frac{2}{(1−2k)}$
$\to - 2x + 1 = −\frac{(3k+4)}{(1−2k)}x+\frac{2}{(1−2k)}$
Assuming we can compare coefficients, we have
$- 2 = −\frac{(3k+4)}{(1−2k)}$
$1 = \frac{2}{(1−2k)}$
giving two values of k.
On the other hand, I am doubtful we can compare coefficients because this time we know x = 1. Plugging in x = 1 into:
$- 2x + 1 = −\frac{(3k+4)}{(1−2k)}x+\frac{2}{(1−2k)}$, we get a unique value of k.
It seems we are allowed to compare coefficients in the partial fraction decomposition problem but not in the quadratic equation problem. Is that right? Why/Why not?
I don't think the problem has to do with comparing coefficients. The problem is that you "say" that the roots are $x_1=x_2=1$. You can't say that because you only have one parameter $k$ and you can't use it to fix both roots. As mathlove points out in a comment, upon substituting $k=-1/5$, the resulting equation doesn't in fact have a double root at $x=1$. Thus none of the conclusions that you draw from this fiat are valid.