I was trying to solving this question:

If roots of the equation $a x^{2}+b x+c=0$ are $\alpha$ and $\beta$, find the equation whose roots are $\frac{1-\alpha}{1+\alpha}, \frac{1-\beta}{1+\beta}$

I was not able to solve it so I looked to the solution given in the book, it was as follows:

Let $\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{1+y}$ .

Now $\alpha$ is root of the equation $a x^{2}+b x+c=0$ $\Rightarrow a \alpha^{2}+b \alpha+c=0$ $\Rightarrow \quad a\left(\frac{1-y}{1+y}\right)^{2}+b\left(\frac{1-y}{1+y}\right)+c=0$ . Hence required equation is $a(1-x)^{2}+b\left(1-x^{2}\right)+c(1+x)^{2}=0$

In the first step the auther used componendo dividend rule as follows:

$\begin{aligned} & \frac{1-\alpha}{1+\alpha}=y \\ \Rightarrow & \frac{1-\alpha+1+\alpha}{1-\alpha-(1+\alpha)}=\frac{y+1}{y-1} \\ \Rightarrow & \frac{2}{-2 \alpha}=\frac{y+1}{y-1} \\ \Rightarrow & \frac{-1}{\alpha}=\frac{y+1}{y-1} \Rightarrow \alpha=\frac{1-y}{y+1} \end{aligned}$

But I don't understand how he could use it, as we are not sure whether $y$ could be - 1 and hence $1 +y$ can be zero. So is this an error? If so then how can we solve this question and if not then why not?

  • 3
    $\begingroup$ Just isolate $y=-1$ as a separate case. Here it is a minor flaw in the book's reasoning. $\endgroup$
    – Trebor
    Jul 10 at 9:35

Notice that $y$ cannot be equal to $-1$ in the first place, if $y = \dfrac{1-\alpha}{1+\alpha}$

Range 0f $f(\alpha) = \dfrac{1-\alpha}{1+\alpha}$ is $\mathbb{R} \setminus \{-1\} ~\forall ~\alpha \in \mathbb{R}$

This is because if $\dfrac{1-\alpha}{1+\alpha} = -1$ then $$ 1-\alpha = -1-\alpha$$ $$ 1=-1$$ Which is clearly meaningless.

In general, $\dfrac{ax+b}{cx+d}$ cannot be equal to $\dfrac{a}{c} $ if $\dfrac{a}{c} \ne \dfrac{b}{d}$

  • 2
    $\begingroup$ The last is only true when $ad-bc\ne 0$ $\endgroup$ Jul 10 at 9:45
  • $\begingroup$ @HagenvonEitzen Yes, indeed. I should have been a bit more careful when making a general comment. $\endgroup$
    – Ankit Saha
    Jul 10 at 9:50

Is $\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{y+1}$ correct even if $y=-1$?

The tag is given. The answer to the question is: No.

But with tag , the answer to the question would be: Yes.
Calculating in the Riemann sphere, we have $$ \frac{1-\infty}{1+\infty} = -1\qquad\text{and}\qquad \frac{1-(-1)}{(-1)+1} = \infty . $$


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