# 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:

‎$1. ‎\theta‎‎‎_{2}‎‎‎‎+‎\theta‎‎_{1}‎<1‎‎$

$2. ‎\theta‎_{2}‎‎-‎\theta‎_{1}<1‎$

$3. ‎-1<‎\theta‎_{2}<1$‎‎‎

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(1) What are $\,\beta\,,\,\theta\,$? (2) In the inequality, is the whole thing divided or only what's in the square root? (3) Please do try to use \frac{.}{.} to denote division, otherwise is hard to tell apart. – DonAntonio Nov 19 '12 at 17:30
Please make your question more clearer by noting what you have done and also by additional tag. For example in which area this lemma is brought up. – Babak S. Nov 19 '12 at 17:31
@DonAntonio: ‎$‎\theta‎_{1}and‎‎ ‎\theta‎_{2} ‎are ‎‎coefficients‎‎$‎ – M.Sina Nov 19 '12 at 18:05

$\frac{\theta_1\pm \sqrt {\theta_1^2+4\theta_2}}{2}$ 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.