Mathematician Finds Easier Way to Solve Quadratic Equations

[carey]

Max, respectfully, your earlier point seemed to be that the quadratic equation is relatively unimportant because, in your opinion, linearity reigns supreme.

(05-05-2024 12:48 PM)Maximilian Hohmann Wrote:  I have never since finishing school come across a quadratic equation in real life. Things are mostly simple and can be handled with linear calculations ("if my plane needs 1400 pounds of fuel in one hour, how much must I take along for a 2:25h flight?") or they are really complicated and require higher order polynomials, transcendental funcions or numeric solutions. But a quadratic equation? Never ever.

Now your point seems to have shifted to the quadratic formula for solving quadratic equations is implemented in software, so why learn it.

(05-05-2024 09:18 PM)Maximilian Hohmann Wrote:  Yes, but for daily work this is done using tables (until 20 years ago) or rather software on a computer (since 20 years) or an iPad App (since 10 years).

My question really remains: Who has ever, in professional or private life, used his memorised "midnight formula"...?

The former is the view that quadratic equations are unimportant, while the latter is the view that the quadratic formula for solving quadratic equations is unimportant.

Respectfully, I find both views mistaken.

1) The number of uses for quadratic equations is so numerous, and in so many fields, that listing them is like listing uses of the Pythagorean theorem. (see EdS2's current thread for a few examples).

2) Besides the obvious that it's good to understand equations even if we use software to implement them, the quadratic formula in symbolic form is essential for determining equilibria and bifurcations in nonlinear ODE's (for which software is limited to numerical solutions since most nonlinear ODEs have no closed-form solutions).

For example, to find equilibria solutions and bifurcations in the nonlinear ODE \[ \frac{dy}{dt} = y^2 +2y + a \] set it equal to 0, resulting in a non-factorable quadratic equation \[y^2 + 2y + a = 0\].
Solving for y in terms of parameter a using the quadratic formula
\[-b \pm \frac{\sqrt {(b^2 - 4ac)}}{2a} \]
\[-2 \pm \frac{\sqrt {(4 - 4a)}}{2} \]
\[-1 \pm \sqrt {(1 - a)} \]
we obtain an expression (1 - a under the radical) that gives the number of equilibrium solutions and shows the importance of the quadratic formula used as a symbolic equation.

Case 1: If 1-a =0, then a = 1 and there is a single equilibrium solution: y=−1.

Case 2: If 1-a>0, then a <1 and there are two equilibrium solutions: $$-1 \pm \sqrt {(1 - a)} $$

Case 3: If 1-a <0, then a>1 and there are no equilibrium solutions since y is complex.

At a = 1, the number of equilibrium solutions changes from 2 to 1 to 0, as the value of parameter a goes from a<1 to 1 to a>1. Therefore a = 1 is a bifurcation point (and is the only bifurcation point since the number of equilibrium solutions doesn't change at any other value of a). For more details about this example, see Section 2.4 Bifurcation in LibreText MTH 225 Differential Equations.

Summarizing, using the quadratic formula in symbolic form gave the equilibria and bifurcation points to a nonlinear ODE -- not bad for a useless formula! :)

Nonlinear ODEs are everyday staples for engineers working with MEMS and NEMS sensors and nonlinear control, for brain neuroscientists, applied mathematicians, and on and on. While all of us can make it through dinner without using the quadratic formula, many would be unable to get through a day without it! :)