Is one of the roots of a linear equation $\infty$?

Question

Consider the quadratic $ax^{2}+bx+c=0$. This equation is linear if $a=0$. Suppose we write $x=\frac{1}{y}$, we get $a+by+cy^{2}=0$. Since $a=0$, we get $y=-\frac{b}{c},0$. Thus, we get two roots of the equation $ax^{2}+bx+c=0$ which are $x=\frac{-c}{b},\infty$. Does this argument mean anything or have any significance in any branch of math or is it just one of the discrepancies of dividing by $0$?

Answer 1

The argument is correct and does have significance given the right context (projective geometry, as suggested in the comments), but there is a subtle point. The question in the title has a negative answer as stated: linear equations still have a single solution in the projective setting. The key point is that in this context, linear equations are not the same as quadratic equations where the first coefficient is zero.

The projective solutions of a polynomial equation $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0$ are by definition the ratios $X:Y$, where $X$ and $Y$ are solutions of the homogeneized equation $a_n X^n + a_{n-1} X^{n-1} Y + \cdots + a_1 X Y^{n-1} + a_0 Y = 0$ (we treat the ratios $X:Y$ and $\lambda X:\lambda Y$ as equivalent if $\lambda$ is any nonzero constant). In particular, given a quadratic equation $a x^2 + b x + c = 0$, its homogeneous counterpart is $a X^2 + b XY + c Y^2$. If $a=0$, this becomes $b XY + c Y^2 = 0$. We can factor this as $Y(bX + cY) = 0$, so we can see that one of the solutions is $X:Y = 1:0$ (which corresponds to the point at infinity, $\infty$), and the other is $X:Y = -c:b$ (which, if $b\neq 0$, represents the finite solution $x=-c/b$).

If we had started with the linear equation $b x + c = 0$ instead, which becomes $bX + cY = 0$ in homogeneous form, the $X:Y = 1:0$ solution is lost, and only the finite solution remains. So, as weird it sounds, in a projective setting there is a difference between a linear equation and a quadratic equation where the first coefficient happens to be equal to zero. Said otherwise, if we want to find projective solutions of a given polynomial equation (in non-homogeneized form), the intended degree of the polynomial must always be specified, precisely to avoid this kind of ambiguity.

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One application of this is the computation of the fixed points of a Möbius transformation in the Riemann sphere, which is the projective line over the complex numbers. For a generic transformation $\frac{ax+b}{cx+d}$, its fixed points are the solutions of the quadratic equation

$$\frac{ax+b}{cx+d}=x \quad \to \quad c x^2+(d-a) x-b=0.$$

When $c=0$, the Möbius transformation is affine (of the form $\frac{a}{d}x + \frac{b}{d}$), and the corresponding quadratic equation becomes linear, with apparently only one solution. But as the previous argument reveals, the missing fixed point must be the point at infinity. One can in fact characterize affine transformations as the Möbius transformations with one fixed point at $\infty$.

Additionally, when $c=0$ and $a=d$ (so that the Möbius tranformation is a simple translation $x \mapsto x+b$), the fixed point equation becomes $b=0$, which apparently has no solutions. In this case, there is a single fixed point at $\infty$ with multiplicity $2$ (Möbius transformations where both fixed points coincide are called parabolic transformations; translations are an example of this).

Answer 2

Division by zero is undefined

When you substitute $x=\frac{1}{y}$, you directly declare a condition that $y$ should not be zero, hence $y=0$ is an extraneous solution and hence it is to be rejected, the other solution is the one and only correct solution.