Quadratic equation and inequality of solution

Solution 1:

Given $x_{+} = \frac{- a +\sqrt{a^2 - b^2}}{b},x_{-} = \frac{- a -\sqrt{a^2 - b^2}}{b}$, the following is true with a>b>0 :

|$x_{-}| = |\frac{ a +\sqrt{a^2 - b^2}}{b}| >|\frac{ a +\sqrt{b^2 - b^2}}{a}=1$ , |$x_{+}| = |\frac{- a +\sqrt{a^2 - b^2}}{b}|<|\frac{- b +\sqrt{a^2 - a^2}}{b}|=1$,

$\to$|$x_{+}|^{2} < 1 < |x_{-}|^2$, $x_{+}^{2} < 1 < x_{-}^2$

Solution 2:

From $bx^2+2ax+b=0$, we can compute the discriminant, of which $4(a^2-b^2)>0$, and we conclude that there are two distinct roots. Furthermore since $a>0$, $b>0$, the roots must be negative.

Hence $x_- < x_+<0$

Also,

$$x^2+\frac{2a}{b}+1=0$$

From the constant term,we can conclude that $x_+x_-=1$, hence $x_- < -1 < x_+<0$.

Hence, we must have $x_+^2 <1 < x_-^2$.