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Given two real quadrics in $\mathbb P^2$ without common components, then we know that they have at most 4 intersection points.

Suppose that the quadrics are not degenerate (I don't know if this is necessary) and that they are in the form

$$P_1(x,y) = x + Q_1(x,y) \qquad P_2(x,y) = y + Q_2(x,y)$$

where $Q_1$ and $Q_2$ are homogeneous of degree 2, and $Q_1(x,y)=Q_2(x,y)=0$ implies $x=y=0$.

Obviously, they have the common solution $(0,0)$ with multiplicity 1, and they do not intersect on the infinite line thanks to the hypothesis on $Q_1,Q_2$.

I know that there is at least an other intersection between the two quadrics, but I don't know how to prove it. Geometrically, I can see it, because the only quadrics are parabolas, circles and hyperboles, and if two of them have a simple intersection, then they have an other one. How can I prove it without doing all the computations case by case?

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    $\langle x+a x^2+b x y+c y^2$, $y+a' x^2+b' x y+c' y^2\rangle$ has a lex grobner basis that includes $$(c^2a'^2-bca'b'+acb'^2+b^2a'c'-2aca'c'-abb'c'+a^2c'^2)y^4+(b^2a'-2aca'-abb'-ca'b'+2a^2c'+2ba'c'-ab'c')y^3+(a^2+2ba'-ab'+a'c')y^2+a'y$$ which must have a real solution besides 0. – 2018-02-25
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    not in the case $a'=0$ or when the coefficient of $y^4$ is zero. – 2018-02-26
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    When $a’=0$, $y+\ldots$ is degenerate (a pair of lines). The coefficient of $y^4$ might have a similar consequence, I’ll investigate. – 2018-02-26
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    The coefficient of $y^4$ simplifies to $(ac'-ca')^2+(bc'-cb')(ba'-ab')$, so you're right, more is needed for the complete result. – 2018-02-26
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    Investigating in geogebra the discriminant of $(b^2a'-2aca'-abb'-ca'b'+2a^2c'+2ba'c'-ab'c')y^2+(a^2+2ba'-ab'+a'c')y+a'$ when $(ac'-ca')^2+(bc'-cb')(ba'-ab')=0$ I found $$\langle x-0.9x^2-3.15xy-2.25y^2,y+5x^2+0xy-5y^2\rangle$$ which seems to be a counter-example. – 2018-03-03
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    But $$-0.9x^2-3.15xy-2.25y^2=5x^2-5y^2=0$$ has a common component $y+x=0$. – 2018-03-04
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    @Jan-MagnusØkland Write a proper answer – 2018-03-05

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What if you have $f(x)=-3x^2$ and $g(x)=\frac{1}{2}x^2$? Then you have a single intersection at $(0,0)$ but no intersections anywhere else.

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    your conics intersect at $(0,y)$ for every $y$.. – 2017-01-13
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    @Exodd, I guess I don't understand the notation in your original question. But your comment doesn't make sense to me; how can the functions intersect at $(0, y)$ for all $y$ when each function only crosses the $x$-axis at $(0, 0)$? I mean, neither function includes the point $(0,1)$. – 2017-01-13
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    Sorry, I'll try to explain: my conics have two variables $x,y$, for example $x^2+2y^2-xy+x$, and I'm looking for the intersections of their zero sets. – 2017-01-13