# 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

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