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The problem is stated below

Find the range of $\alpha$ such that $1-x-\alpha x^2$ has roots greater than $1$ or less than $-1$ (i.e if $\mu_1$ and $\mu_2$ are the roots of $1-x-\alpha x^2$, we require $|\mu_1|, |\mu_2|>1$).

Solution:

Transform the above question into the following. We require the roots of the quadratic equaion $\frac{1}{x^2}-\frac{1}{x}-\alpha$ have absolute value smaller than 1. Its roots are given by $$\mu_1=\frac{1+\sqrt{1+4\alpha}}{2}, \mu_2=\frac{1-\sqrt{1+4\alpha}}{2}$$

Case 1: If $1+4\alpha > 0$, then $|\mu_1|<1$ implies $\alpha <0$.

Case 2: If $1+4\alpha < 0$, then $|\mu_1|<1$ is equivalent to $(\frac{1}{2})^2-(\frac{\sqrt{1+4\alpha}}{2})^2<1$ which implies $\alpha > -1$.

Thus the solution is given by $-1<\alpha<0$.

My Questions:

  1. In the solution given, only the absolute value of $\mu_1$ is taken into consideration. Why?

  2. Also, I don't understand how $(\frac{1}{2})^2-(\frac{\sqrt{1+4\alpha}}{2})^2<1$ is come from. I know that when $1+4\alpha < 0$, the roots would become complex number. Yet, I still can't come up with the expression above.

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  1. In the solution given, only the absolute value of $μ_1$ is taken into consideration. Why?

Because, as noted just above that: "We require the roots ... have absolute value smaller than 1".

  1. Also, I don't understand how $(\frac{1}{2})^2-(\frac{\sqrt{1+4\alpha}}{2})^2<1$ is come from. I know that when $1+4\alpha < 0$, the roots would become complex number.

When $1+4 \alpha \lt 0$ let $\beta = -(1+4 \alpha) \gt 0$, so that $\sqrt{\beta}= i\, \sqrt{1+4 \alpha}$. Then the roots can be written as $\mu_{1,2} = \frac{1}{2} \pm i \frac{\sqrt\beta}{2}$ and $|\mu_{1,2}|^2 = \left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{\beta}}{2}\right)^2 = \left(\frac{1}{2}\right)^2+\left(\frac{i \; \sqrt{1 + 4 \alpha}}{2}\right)^2 = \left(\frac{1}{2}\right)^2-\left(\frac{\sqrt{1 + 4 \alpha}}{2}\right)^2$.


[ EDIT ]    To answer the question from the comment, in Case 1:

$$\mu_1=\frac{1}{2} + \frac{\sqrt{1+4\alpha}}{2} \lt 1 \iff \frac{\sqrt{1+4\alpha}}{2} \lt \frac{1}{2} \iff \sqrt{1+4\alpha} \lt 1 \iff \alpha \lt 0$$

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  • $\begingroup$ So you mean for case 1 we have $|\mu_2|=l\mu_1l=\frac{1+\sqrt{1+4\alpha}}{2}$? I can't see why... $\endgroup$ – IDontKnowMath Nov 2 '16 at 4:52
  • $\begingroup$ see my addition... $\endgroup$ – Momo Nov 2 '16 at 4:59
  • $\begingroup$ @IDontKnowMath I did not mean, and did not write that $|\mu_1|=|\mu_2|$ in Case 1. As to why $|\mu_1| = \mu_1 \lt 1$ implies $\alpha \lt 0$ see the edit at the end of the answer above. $\endgroup$ – dxiv Nov 2 '16 at 5:04
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For the first question:

Case 1. if $\alpha < 0$ then $0<\mu_2<1/2$ because the second root is 1/2 * (1 - something positive less than 1), so the second root is automatically less than 1

Case 2. You have complex roots, the absolute value (magnitude) of the roots is the same, so if $|\mu_1|<1$ then $|\mu_2|=|\mu_1|<1$

I would also add that Case 1 should be $1+4\alpha \ge 0$, so that $\alpha=-1/4$ is included in one of the cases :)

The rest has been already answered by divx

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  • $\begingroup$ Thank you for the answer! That's really what I was struggling when reading the solution. $\endgroup$ – IDontKnowMath Nov 2 '16 at 5:27

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