# 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.

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

• My answer is more complicated because we need to find values of polinomial zeros by approximation and we can't just use Vietta's formula because it not giving us concrete x_1 and x_2. May 16 at 17:18

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

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

• Thank you very much, I didn't expect, that it would be that easy. May 16 at 16:40