# 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 :

$$x^2=x+y+4 , y^2=y-15x+36$$

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Looks like you need to solve the system. Try to find $y$ in terms of $x$ from the first equation, and plug into the second one. –  gt6989b Sep 23 '12 at 19:15
Set $y=0$, solve both for $x$, add the results. (Assuming I interpreted your question correctly.) –  Alex Becker Sep 23 '12 at 19:16

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 $-(-6)$.

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Is the second curve a function ? –  Frank Sep 28 '12 at 13:42
@MohammedAl-mubark: In the second curve, $y$ is not a function of $x$, but $x$ is a function of $y$. –  André Nicolas Sep 28 '12 at 13:55
Ok ,but is there any relationship between the two curves if we draw them on the plane ? –  Frank Sep 28 '12 at 17:33
They are both parabolas, maybe even closely related parabolas. –  André Nicolas Sep 28 '12 at 17:35