Circuit Theory/Quadratic Equation Revisited

The Quadratic Equation
The roots to the quadratic polynomial

$$ y = ax^2 + bx +c $$

are easily derived and many people memorized them in high school:

$$ \begin{align} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \end{align} $$

Derivation of the Quadratic Equation
To derive this set $$y = 0 $$ and complete the square:

$$ \begin{align} 0 &= ax^2 + bx +c \\ 0 &= a \left( x^2 + \frac{b}{a} x + \frac{c}{a} \right)\\ 0 &= x^2 + \frac{b}{a} x + \left(\frac{b}{2a}\right)^2 + \frac{c}{a} - \left(\frac{b}{2a}\right)^2 \\ 0 &= \left( x + \frac{b}{2a} \right)^2 + \left(\frac{c}{a} - \left(\frac{b}{2a}\right)^2 \right). \end{align} $$

Solving for $$x$$ gives

$$ \begin{align} 0 &= \left( x + \frac{b}{2a} \right)^2 + \left(\frac{c}{a} - \left(\frac{b}{2a}\right)^2\right) \\ \left( x + \frac{b}{2a} \right)^2 &= -\left(\frac{c}{a} - \left(\frac{b}{2a}\right)^2\right) \\ &= \frac{b^2 - 4ac }{\left(2a\right)^2}. \end{align} $$

Taking the square root of both sides and putting everything over a common denominator gives

$$ \begin{align} x + \frac{b}{2a} &= \pm \sqrt{\frac{b^2 - 4ac }{\left(2a\right)^2}} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \end{align} $$

Numerical Instability of the Usual Formulation of the Quadratic Equation
Middlebrook has pointed out that this is a poor expression from a numerical point of view for certain values of $$a$$, $$b$$, and $$c$$.

[Give an example here]

Middlebrook showed how a better expression can be obtained as follows. First, factor $$-\frac{b}{2a}$$ out of the expression:

$$ \begin{align} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \left( -\frac{b}{2a} \right) \left( 1 \pm \sqrt{1 - \frac{4ac}{b^2}} \right). \end{align} $$

Now let

$$ Q^2 = \frac{ac}{b^2}. $$

Then

$$ \begin{align} x &= \left( -\frac{b}{2a} \right) \left( 1 \pm \sqrt{1 - 4Q^2} \right) \end{align} $$

A More Numerically Stable Formulation of the Negative Root
Considering just the negative square root we have

$$ \begin{align} x_1 &= \left( -\frac{b}{2a} \right) \left( 1 - \sqrt{1 - 4Q^2} \right). \end{align} $$

Multiplying the numerator and denominator by $$1 + \sqrt{1 - 4Q^2}$$ gives

$$ \begin{align} x_1 &= \left( -\frac{b}{2a} \right) \left( 1 - \sqrt{1 - 4Q^2} \right) \left( \frac{1 + \sqrt{1 - 4Q^2}}{1 + \sqrt{1 - 4Q^2}} \right) \\ &= \left( -\frac{b}{2a} \right) \left( \frac{1 - 1 + 4Q^2}{1 + \sqrt{1 - 4Q^2}} \right) \\ &= -\frac{2 b Q^2}{a} \left( \frac{1}{1 + \sqrt{1 - 4Q^2}} \right) \\ &= -\frac{2 b \left( \frac{ac}{b^2} \right)}{a} \left( \frac{1}{1 + \sqrt{1 - 4Q^2}} \right) \\ &= -\frac{2 c}{b} \left( \frac{1}{1 + \sqrt{1 - 4Q^2}} \right) \\ &= -\frac{c}{b} \left( \frac{1}{\frac{1}{2} + \frac{1}{2}\sqrt{1 - 4Q^2}} \right). \end{align} $$

By defining

$$ F = \frac{1}{2} + \frac{1}{2}\sqrt{1 - 4Q^2} $$

we can write

$$ \begin{align} x_1 &= -\frac{c}{bF}. \end{align} $$

Note that as $$Q \to 0$$, $$F \to 1$$.

Finding the Positive Root Using the Same Approach
Turning now to the positive square root we have

$$ \begin{align} x_2 &= \left( -\frac{b}{2a} \right) \left( 1 + \sqrt{1 - 4Q^2} \right) \\ &= \left( -\frac{b}{a} \right) \left( \frac{1}{2} + \frac{1}{2}\sqrt{1 - 4Q^2} \right) \\ &= -\frac{bF}{a}. \end{align} $$

Using the two roots $$x_1$$ and $$x_2$$, we can factor the quadratic equation

$$ \begin{align} y &= a x^2 + b x + c \\ &= a \left( x^2 + \frac{b}{a} + \frac{c}{a}\right) \\ &= a \left( x - x_1\right) \left( x - x_2\right) \\ &= a \left( x + \frac{c}{bF}\right) \left( x +\frac{bF}{a}\right). \end{align} $$

Accuracy for Low $$Q$$
For values of $$Q \le 0.3$$ the value of $$F$$ is within 10% of 1 and we may neglect it. As noted above, the approximation gets better as $$Q \to 0$$. With this approximation the quadratic equation has a very simple factorization:

$$ \begin{align} y &= a x^2 + b x + c \\ &\approx a \left( x + \frac{c}{b}\right) \left( x +\frac{b}{a}\right), \end{align} $$

an expression that involves no messy square roots and can be written by inspection. Of course, it is necessary to check the assumption about $$Q$$ being small before using the simplification. Without this simplification, $$F$$ needs to be calculated and the roots are slightly more complicated.

[Explore the consequences if $$Q > 0.5 $$.]