Quadratic equation and inequality of solution

Question

I have a quadratic equation, $bx^2 + 2ax + b =0$, with, $a>b>0$. I can solve this as followings, $x_{+} = \frac{- a +\sqrt{a^2 - b^2}}{b},x_{-} = \frac{- a -\sqrt{a^2 - b^2}}{b},$

In my text book, an inequality equation, $x_{+}^{2} < 1 < x_{-}^2$, is written with the solution, but I cannot derive the equation. If you can, please teach me the derivation. Best regards,

Answer 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 :

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

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

Answer 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 $x_+^2 <1 < x_-^2$.

Answer 3

You have expressions for the zeroes $ \ x_{+} \ = \ \frac{- a +\sqrt{a^2 - b^2}}{b} \ , \ x_{-} = \frac{- a -\sqrt{a^2 - b^2}}{b} \ \ , $ with $ \ a \ > \ b \ > \ 0 \ \ . $ So we can first determine that $$ \frac{ \sqrt{a^2 \ - \ b^2}}{b} \ \ = \ \ \frac{ a· \sqrt{1 \ - \ \left(\frac{b}{a} \right)^2}}{b} \ \ < \ \ \frac{a}{b} \ \ , $$ and thus, $$ - \ \frac{ a· \sqrt{1 \ - \ \left(\frac{b}{a} \right)^2}}{b} \ \ < \ \ 0 \ \ < \ \ \frac{ a· \sqrt{1 \ - \ \left(\frac{b}{a} \right)^2}}{b} \ \ < \ \ \frac{a}{b} $$ $$\Rightarrow \ \ x_{-} \ = \ -\frac{a}{b} \ - \ \frac{ a· \sqrt{1 \ - \ \left(\frac{b}{a} \right)^2}}{b} \ \ < \ \ -\frac{a}{b} \ \ < \ \ x_{+} \ = \ -\frac{a}{b} \ - \ \frac{ a· \sqrt{1 \ - \ \left(\frac{b}{a} \right)^2}}{b} \ \ < \ \ 0 \ \ . $$

A quadratic polynomial of the form $ \ A·x^2 + B·x + A \ $ is called palindromic, a characteristic of which is that if $ \ r \ $ is a zero, then $ \ \frac{1}{r} \ $ is also. (A proof of this is suggested by the second paragraph of Siong Thye Goh's answer.) So $ \ x_{-} \ = \ \frac{1}{x_{+}} \ \ . $ We can also demonstrate this by direct calculation: $$ \frac{1}{x_{+}} \ \ = \ \ \frac{b}{- a \ + \ \sqrt{a^2 - b^2}} \ \ = \ \ \frac{b · ( \ - a \ - \ \sqrt{a^2 - b^2} \ )}{( \ - a \ + \ \sqrt{a^2 - b^2}\ ) \ · \ ( \ - a \ - \ \sqrt{a^2 - b^2} \ )} $$ $$ = \ \ \frac{b · ( \ - a \ - \ \sqrt{a^2 - b^2} \ )}{ a^2 \ - \ (a^2 - b^2 )} \ \ = \ \ \frac{ - a \ - \ \sqrt{a^2 - b^2} }{ b} \ \ = \ \ x_{-} \ \ . $$ We find (from Viete, for instance) that the sum of the roots is $ \ x_{+} + x_{-} \ = \ -\frac{2a}{b} \ < \ -2 \ \ . $ It follows that $$ \left( x_{+} \ + \ \frac{1}{x_{+}} \right)^2 \ \ = \ \ x^2_{+} \ + \ 2 \ + \ \frac{1}{x^2_{+}} \ \ > \ \ 4 \ \ \Rightarrow \ \ x^2_{+} \ + \ \frac{1}{x^2_{+}} \ \ > \ \ 2 \ \ . $$ From the AM-GM inequality and many other starting points, we can show that $ \ a \ + \ \frac{1}{a} \ \ge \ 2 \ \ $ for $ \ a > 0 \ \ , $ with equality holding for $ \ a = 1 \ \ . $ Our result indicates then that the larger of $ \ x^2_{+} \ $ or $ \ \frac{1}{x^2_{+}} \ $ must be greater than $ \ 1 \ \ . $ Since we have $ \ x_{-} \ < \ x_{+} \ < \ 0 \ \ , $ we conclude that $ \ x^2_{-} \ $ is the larger of the two and hence $ \ x^2_{+} \ < \ 1 \ < \ x^2_{-} \ \ . $