Question regarding an algorithm for solving quadratic equations I'm reading currently Martin Hanke Burgeois' Book about numeric analysis. It is in German, but my question is the following and I think it's not that hard:
In a subsection dedicated to finding zeros of quadratic polynomials of the form $ax^2 + bx + c$ with $a,c \neq 0$ and $b^2 - 4ac > 0$ he argues, that there can be two cases, where the problem can have a high condition number, if we apply the formula:
$x_{1/2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, namely:
If $b^2 \gg |4ac|$ and $b>0$ he proposes as a possibility, to approximate $x_2$ by the above formula with the $-$ sign and for $x_1$ to use $x_1 = \frac{c}{ax_2}$.
I would be very happy, if someone could tell me, why this would work and where does this idea come from.
Thanks to all in advance
 A: Let $ x_1 $ and $ x_2 $ be the real roots of the equation
$$ax^2+bx+c=0$$
with the conditions
$$a\ne 0\text{ and } b^2-4ac>0$$
in other words, we have
$$x_1=\frac{-b-\sqrt{b^2-4ac}}{2a}$$
and
$$x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}$$
their product is
$$x_1.x_2=\frac{b^2-b^2+4ac}{4a^2}=\frac ca$$
thus
$$x_1=\frac{c}{ax_2}$$
A: Let's take attention to this proposition: $ b^2≫|4ac| $.
Symbol $ ≫ $ means that left term is significantly larger than right term, not just larger $ > $.
From $ b^2≫|4ac| $ we can conclude that discriminant $ b^2 - 4ac $ is almost equal to $ b^2 $ since square root of discriminant $ \sqrt{b^2 - 4ac} $ is almost equal to $ b $.
More detailed view:
Assumption:

*

*$ b^2 - |4ac| ≫ 0 $
Can be rewritten as two cases (remember $ ac \ne 0$ ):

*

*if $ ac > 0$ then $ b^2 - 4ac ≫ 0 $

*if $ ac < 0$ then $ b^2 + 4ac ≫ 0 $
We can glue this two cases together because if $ b^2 - |4ac| ≫ 0 $ then $ |b| ≫ 0 $.
And we need just approximation for zeros $ x_{1,2} $ of polinomial.
At this point we know that $ \sqrt{b^2 - 4ac} $ is almost equal to $ b $.
But now we have two ways:

*

*First thinking about $ b $ as very large number and is hard to check that all formulas is right.

*Second way is introducing some variable $ \Delta $ that denotes approximation error.

If we select second then we can rewrite "$ \sqrt{b^2 - 4ac} $ is almost equal to $ b $" as $ \sqrt{b^2 - 4ac} = b + \Delta $ where $ |b| ≫ |\Delta | $.
Let's substitute this into equation for $ x_2 $:
$$ x_{2} = \frac{-b-\sqrt{b^2 - 4ac}}{2a} $$
after substitution we got
$$ x_{2} = -\frac{b + \Delta}{a} $$
But $ |b| ≫ |\Delta | $, and relative error for $ b $ is $ \epsilon = \frac{b + \Delta}{b} $ is almost equal to $ 1 $ and we can safely replace $ b + \Delta $ with $ b $:
$$ x_{2} = -\frac{b}{a} $$
Ok, we have very short way to find value of $ x_2 $, but after same step for $ x_1 $ we got something strange
$$ x_{1} = \frac{-b+b+ \Delta}{2a} $$
simplifies to (remember that $ a \ne 0 $ that follows from $ ac \ne 0 $)
$$ x_{1} = \frac{\Delta}{2a} $$
This follows that $ \Delta $ have powerful impact on $ x_1 $ value and $ x_1 $ can have approximation error even larger than $ x_2 $.
And of course for $ x_1 $ we need to find formula with less approx error.
That's why we firstly must find $ x_2 $ and after $ x_1 $ somehow ($ - $sign case).
From Vieta's formulas for quadratic polinomial $a x^2 + b x + c$ we know if $ b^2 - 4ac \gt 0 $:
$$ x_1 x_2 = \frac{c}{a} $$
$$ x_1 + x_2 = -\frac{b}{a} $$
If we use second formula we again have large error and this is not we want.
Let's use first $ x_1 x_2 = \frac{c}{a} $:
$$ x_1 = \frac{c}{a x_2}$$
And this formula have indeed less approximation error because (we can rewrite it to make $ \Delta $ visible using $ x_{2} = -\frac{b + \Delta}{a} $):
$$ x_1 = -\frac{c}{a} \frac{a}{b + \Delta} $$
after simplification
$$ x_1 = -\frac{c}{b + \Delta} $$
and after using $ |b| ≫ |\Delta | $ we got
$$ x_1 = -\frac{c}{b} $$
