Relationship Between Roots and Coefficients of a Quadratic To prove this lemma I use  the relationship between roots and coefficients of a quadratic equation but did not get the result.
Please help me prove this lemma.
If ‎‎  $ - ‎\theta‎‎_{2}x^2 - ‎ \theta‎‎_{1}‎x + 1 = 0‎‎$‎‎‎ and 
 ‎$‎\left| ‎‎\frac{ \theta‎‎‎_{1}‎ ‎\mp ‎\sqrt{(‎\theta‎‎_{1}^2+4‎\theta‎‎_{2})‎}}{-‎2‎\theta‎‎‎_{2}} ‎\right|‎‎‎< 1$‎ then:
‎$1. ‎\theta‎‎‎_{2}‎‎‎‎+‎\theta‎‎_{1}‎<1‎‎$
$2. ‎\theta‎_{2}‎‎-‎\theta‎_{1}<1‎$
$3.  ‎-1<‎\theta‎_{2}<1$‎‎‎
 A: $\frac{\theta_1\pm \sqrt {\theta_1^2+4\theta_2}}{2}$ are the two roots of $z^2-\theta_1 z -\theta_2=0$, which are the reciprocals of the roots of $1-\theta_1 x-\theta_2x^2=0$.
Thus the premise is that the roots of $z^2-\theta_1 z -\theta_2$ are smaller in absolute value than ${\theta_2}$.
Since $\theta_2$ is also the (negative) product of the roots, each root is $>1$ in absolute value.
Note that 
$\frac{\theta_1+ \sqrt {\theta_1^2+4\theta_2}}{-2\theta_2}\cdot \frac{\theta_1- \sqrt {\theta_1^2+4\theta_2}}{-2\theta_2} = -\frac1{\theta_2}$.
Since the numbers on the left are $<1$ in absolute value, we conclude that $|\theta_2|>1$, hence claim 3 is clearly wrong.
The numbers $\frac{\theta_1\pm \sqrt {\theta_1^2+4\theta_2}}{2}$ are also the roots of $f(z)=z^2-\theta_1 z -\theta_2$.
Claim 1 is that $f(1)>0$, claim 2 is that $f(-1)>0$.
Note that $f$ is a parabola oben upwards and  assumes its minimum on the real line at $z=\frac{\theta_1}2$, with value $-\frac{\theta_1^2+4\theta_2}{4}$. If this number is positive, then even more so must we have $f(\pm1)>0$.
And if $-\frac{\theta_1^2+4\theta_2}{4}$ is nonpositive, then the roots of $f$ are real and the absolute value of their  product equals $|\theta_2|>1$, hence at least one of the roots is $>1$ in absolute value; this seems to work agains the claim that $f(\pm1)>0$. Indeed, you should easily be able to construct counterexamples based on the above calculations.
