Circuit Theory/Quadratic Equation Revisited

The Quadratic Equation

The roots to the quadratic polynomial

$y=a{x}^2+bx+c$

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

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

Derivation of the Quadratic Equation

To derive this set $y=0$ and complete the square:

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

Solving for $x$ gives

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

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

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

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:

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

Now let

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

Then

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

A More Numerically Stable Formulation of the Negative Root

Considering just the negative square root we have

$\begin{array}{cc}{x}_1& =(-\frac{b}{2a})(1-\sqrt{1-4Q^2}).\end{array}$

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

$\begin{array}{cc}{x}_1& =(-\frac{b}{2a})(1-\sqrt{1-4Q^2})\left(\frac{1+\sqrt{1-4Q^2}}{1+\sqrt{1-4Q^2}}\right)\\ & =(-\frac{b}{2a})\left(\frac{1-1+4Q^2}{1+\sqrt{1-4Q^2}}\right)\\ & =-\frac{2b{Q}^2}{a}\left(\frac{1}{1+\sqrt{1-4Q^2}}\right)\\ & =-\frac{2b\left(\frac{ac}{b^2}\right)}{a}\left(\frac{1}{1+\sqrt{1-4Q^2}}\right)\\ & =-\frac{2c}{b}\left(\frac{1}{1+\sqrt{1-4Q^2}}\right)\\ & =-\frac{c}{b}\left(\frac{1}{\frac{1}{2}+\frac{1}{2}\sqrt{1-4Q^2}}\right).\end{array}$

By defining

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

we can write

$\begin{array}{cc}{x}_1& =-\frac{c}{bF}.\end{array}$

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

$\begin{array}{cc}{x}_2& =(-\frac{b}{2a})(1+\sqrt{1-4Q^2})\\ & =(-\frac{b}{a})(\frac{1}{2}+\frac{1}{2}\sqrt{1-4Q^2})\\ & =-\frac{bF}{a}.\end{array}$

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

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

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:

$\begin{array}{cc}y& =a{x}^2+bx+c\\ & \approx a(x+\frac{c}{b})(x+\frac{b}{a}),\end{array}$

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 $Q>0.5$.]