The sum of the different points on $x$-axis What is the sum of the different points on $x$-axis  and that intersect with the two curves :
$$x^2=x+y+4 , y^2=y-15x+36$$
 A: As the comments have said, we set $y=0$. For the curve $y^2=-15x+36$, we get $x=36/15$. For the curve $x^2=x+y+4$ we get $x^2-x-4=0$. 
By the Rational Roots Theorem, the only conceivable rational roots of $x^2-x-4=0$ are integers that divide $4$. None of these is in fact a root, but that is irrelevant,  the point is that $36/15$ cannot be a root of $x^2-x-4=0$.
Now we still need the sum of the (real) roots of $x^2-x-4=0$. It is easy to see that there are in fact two real roots. We could compute them, using the Quadratic Formula, and then add. But in general the sum of the roots of $x^2+ax+b=0$ is $-a$. So the sum of the roots of $x^2-x-4=0$ is $1$. Add that to $36/15$.
Remark: Consider the polynomial equation $a_0x^n+a_1x^{n-1}+\cdots+a_n=0$, where $a_0\ne 0$. Then the sum of all the roots of the equation in the complex numbers is $-a_1/a_0$. In this formula, a root of multiplicity greater than $1$ counts as many times as it occurs. For example, $x^2-6x+9=0$ as a "double root" at $x=3$. The formula predicts, correctly, that the sum of the roots is $-(-6)$.  
