Would an equation in the form ax + b = c/x be considered quadratic?

Question

For example is $2x + 3 = \frac 5x$ quadratic?

On the one hand it has two solutions, $x = 1$ and $x = -5/2$ which is the number of solutions we'd expect from the fundamental theorem of algebra but on the other hand any equation in the form $ax + b = \frac cx$ would be undefined at $x = 0$ and every quadratic equation I'm familiar with is continuous over all real numbers. Is being continuous over all real numbers necessary for an equation to be called quadratic?

Answer 1

The fact that it has two solutions is not relevant here. The equation $2^x-x=0$ also has two solutions, but nobody would say that it is a quadratic equation. And the equation $x^2+1=0$ is a quadratic equation, in spite of the fact that it has no (real) solution.

A reasonable definition of quadratic equation would be: an equation of the type $q(x)=0$, where $q$ is a polynomial function with degree $2$. It would still be a quadratic equation even if we were only interested in solutions within a certain subset of $\Bbb R$ (such as, say, $(-\infty,1]$ or $\Bbb Q$).

Under this definition, your equation is not a quadratic equation. However, its solutions are the solutions of the quadratic equation $2x^2+3x-5=0$.

Answer 2

This equation is not quadratic owing to its form. However, if the domain is restricted to non-zero numbers only, this could be thought of as a part of a quadratic curve, because it could be reduced to the quadratic form.

Answer 3

Ys it is quadratic because $$2x + 3 = \frac {5}{x}\ \iff 2x^2+3x=5 \\\implies 2z^2 +3x-5=0$$