# Solution of a quadratic equation to satisfy a constraint

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

$$$$0 < 1 - 4\theta^2 < 1 \implies 0 < \sqrt{1 - 4\theta^2} < 1.$$$$

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

$$$$1 - \sqrt{1-4\theta^2} - 2|\theta|\sqrt{1-4\theta^2} > 0$$$$

$$$$\frac{1 - \sqrt{1-4\theta^2}}{2|\theta|} > \sqrt{1-4\theta^2}$$$$

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?

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:$$x_1\cdot x_2 =1\implies |x_2| = {1\over |x_1|} <1$$
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 $$\,(0,1)\,$$.