Solution of a quadratic equation to satisfy a constraint

Question

I have the roots of a quadratic equation as

$$x = \frac{1 \pm \sqrt{1-4\theta^2}}{2\theta}$$ I know that $|\theta| < 1/2$. Among these two roots, I want to find the one which has a value $|x| < 1$.

My attempt is:

It can't be $\frac{1 + \sqrt{1-4\theta^2}}{2\theta}$ as numerator exceeds one and denominator $2|\theta| < 1$, hence the modulus of the root is greater than one.

If $x = \frac{1 - \sqrt{1-4\theta^2}}{2\theta}$:

\begin{equation} 0 < 1 - 4\theta^2 < 1 \implies 0 < \sqrt{1 - 4\theta^2} < 1. \end{equation}

\begin{equation} \left[(1 - \sqrt{1 - 4\theta^2}) - 2|\theta|\right]^2 > 0 \end{equation}

\begin{equation} 1 - \sqrt{1-4\theta^2} - 2|\theta|\sqrt{1-4\theta^2} > 0 \end{equation}

\begin{equation} \frac{1 - \sqrt{1-4\theta^2}}{2|\theta|} > \sqrt{1-4\theta^2} \end{equation}

Using the fact that $0 < \sqrt{1 - 4\theta^2} < 1$, left hand side of the inequality has to be less than one.

Is there a more elegant way of finding this?

Answer 1

You can use Vieta's formula. We can easly see that $x$ is a solution to $$\theta x^2-x+\theta=0$$ Since you figer out $|x_1|>1$ you can use now:$$x_1\cdot x_2 =1\implies |x_2| = {1\over |x_1|} <1$$

Answer 2

Alt. hint: $\;x_1+x_2=1/\theta \gt 0\,$ and $\,x_1x_2=1 \gt 0\,$, so both $\,x_1,x_2 \gt 0\,$. Since $\,x_1x_2=1\,$ it follows that $\,0 \lt x_1 \lt 1 \lt x_2\,$, so the smaller root, corresponding to the "$-\sqrt{\Delta}$" sign, is within $\,(0,1)\,$.