关于三次方程$x^3+d=bx^2$（$b,d > 0$）的圆锥曲线相交几何解法问询

Hey there! Great question—geometric solutions to cubic equations using conic sections are such a neat nod to classical algebraic geometry, so let's walk through this clearly, including the general approach you're curious about.

First, Rewrite the Cubic for Clarity #

Your equation is $x^3 + d = bx^2$ with $b,d>0$. Let's rearrange it to $x^3 - bx^2 + d = 0$—this makes it easier to map to conic intersections.

Option 1: Parabola + Circle (Matching Your Teacher's Example) #

Your teacher used a parabola and circle for $x^3 + ax = d$, so we can replicate that structure here. Here's exactly how to derive the circle step by step:

1. Start with the simple parabola $y = x^2$—this is the go-to for cubics because it turns $x^3$ into $x \cdot y$, a product of linear and quadratic terms that's easy to work with.
2. Substitute $y = x^2$ into your cubic:
   $$x \cdot y - b y + d = 0$$
   This is a linear equation (a degenerate conic). To get a proper circle, we'll build a quadratic equation that, when combined with $y=x^2$, produces your cubic plus one harmless extra root (we can ignore it later).
3. Use the general circle equation: $x^2 + y^2 + Cx + Dy + E = 0$. Substitute $y=x^2$ to get a quartic (4th-degree) equation:
   $$x^4 + (1+D)x^2 + Cx + E = 0$$
4. We want this quartic to match $(x + b)(x^3 - bx^2 + d) = 0$—the extra root $x=-b$ isn't a solution to your cubic (unless $d=2b^3$, which is a rare edge case), so it won't interfere. Expand this quartic:
   $$x^4 - b^2x2 + dx + bd = 0$$
5. Match coefficients with the circle-derived quartic:
   • $1+D = -b^2$ → $D = -b^2 -1$
   • $C = d$
   • $E = bd$
6. Plug these back into the circle equation, and you get the final circle:
   $$x^2 + y^2 + dx - (b^2 + 1)y + bd = 0$$

Now, the $x$-coordinates of the intersections between $y=x^2$ and this circle are exactly the solutions to your cubic—just ignore the $x=-b$ root if it appears.

Option 2: Parabola + Hyperbola (No Extra Roots) #

If you want to avoid dealing with an extra root, you can use a hyperbola instead:

1. Start again with the parabola $y = x^2$.
2. Rearrange your original cubic to isolate $x^2$:
   $$x^2 = \frac{d}{b - x}$$
3. Substitute $y=x^2$, so the hyperbola equation is:
   $$y = \frac{d}{b - x}$$

This hyperbola and parabola intersect exactly at the solutions to your cubic—no extra roots, since $x=b$ can't be a solution (plugging $x=b$ into the cubic gives $b^3 + d = b^3$, which is impossible because $d>0$).

General Approach for Any Cubic #

The core logic works for any cubic equation, and it boils down to translating a cubic into a system of two quadratic equations (conic sections):

• Pick a simple conic as a "bridge"—$y=x^2$ is almost always the easiest, since it converts $x^3$ to $x \cdot y$ and $x^4$ to $y^2$.
• Substitute this conic into your cubic to get a relationship between $x$ and $y$. If that relationship is linear, you can turn it into a non-degenerate conic (like a circle) by multiplying through by a term that introduces a harmless extra root.
• For cubics with different forms (like the $x^3 + ax = d$ your teacher showed), the same steps apply: substitute $y=x^2$, rearrange to get a quadratic equation (like a circle), then solve for intersections.

Quick Test Example #

Let's use $b=2$, $d=3$ (cubic: $x^3 -2x^2 +3=0$):

• The circle equation becomes $x^2 + y^2 +3x -5y +6=0$.
• Intersecting with $y=x^2$ gives the quartic $x^4 -5x^2 +3x +6=0$, which factors to $(x+1)(x^3 -2x^2 +3)=0$. The $x=-1$ root is extra, and the other roots are exactly the solutions to our cubic.

备注：内容来源于stack exchange，提问作者user19170731