Question regarding an algorithm for solving quadratic equations

Question

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

Answer 1

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 $$ x_1 = -\frac{c}{b} $$

Answer 2

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

$$x_1=\frac{c}{ax_2}$$