Algebra , Polynomials

Attacking with limited algebra knowledge but patience the equations eventually give away "weak points".

So ... we're looking for all the $x$ that make both equations true.

$x^2 = x+y+4 \text{ and } y^2 = y -15x+36$ .

One way to start is to isolate $y$ from the first equation and substitute into the second:

$\left( x^2-x-4 \right)^2 = x^2-x-4 -15x+36$

Simplifies to

$x^4-2x^3-8x^2+24x-16 = 0$

The last three terms seem easily factorable. Let's try if that takes us somewhere.

$x^3(x-2)-8(x-2)(x-1) = 0$ $(x-2)\left[ x^3-8(x-1) \right] = 0$ That's really a step forward; $x = 2$ is one of our $x$ -coordinates! $(x-2)\left[ x^3-8(x-1) \right] = 0$ Now we "just need" to know what makes that second factor equal to zero.

$x^3-8(x-1) = x^3-8x+8 = 0$ Eyeballing, or checking low-hanging fruits by inserting integers that intuitively seem to fit. We find that $x = 2$ is a solution!

We thus have $x^3-8x+8 = (x-2)z = 0$ , but we don't know $z$ . Long division tells us $z = x^2+2x-4$ .

We then have
$(x-2)^2\left[ x^2+2x-4 \right] = 0$

Now we just need the $x$ that makes $x^2+2x-4=0$ true.

The classical quadratic formula can be used: $x = \frac{-b \pm \sqrt{b^2-4ac} }{2a} = \frac{-2 \pm \sqrt{(-2)^2-4\cdot1\cdot(-4)} }{2\cdot 1} = -1 \pm \sqrt{5}$

The $x$ -coordinates thus are: $\{-1 + \sqrt{5},-1 - \sqrt{5}, 2\}$ and their sum is $0$ .

We are asked to find the sum of the x-coordinates for the points of intersection of the two curves $x^2 = x + y +4$ and $y^2 = y - 15x + 36$ . Re-arranging terms, the two parabolas are the curves $y=x^2-x-4$ and $15x=-y^2+y+36$ . One can solve this algebraically, using the first equation to substitute for $y$ in terms of $x$ in the second equation.
Use substitutions: $y=x^2 - x -4$ and $y^2 =x^4 -2x^3-7x^2 + 8x+16$ After rearranging, we get the fourth-degree polynomial: $x^4-2x^3-8x^2+24x - 16=0$ . At this point we might recall that the coefficient of the $x^3$ term is the negative of the sum of the roots. But we need to be careful: this sum would double-count repeated roots. There is no escaping the need to factor it. Without getting too much into the process, the rational root theorem implies that any root would have to be a power of 2. Indeed, 2 is itself a repeated root. The quartic factors as $x^4 - 2x^3 -8x^2 + 24x - 16 = (x-2)^2(x^2+2x-4)$ It's clear that the discriminant of $x^2 + 2x - 4$ is non-zero, so this part will have distinct roots (which turn out to be $-1 \pm \sqrt{5})$ . Thus the sum of the x-coordinates of the points of intersection is $2 + (-1 + \sqrt{5}) + (-1 - \sqrt{5}) = 0.$ Alternatively, one can note that the only repeated root is $2$ , so subtracting $2$ from the negative of the original coefficicent of $x^3$ will also give the same solution of $0$ .

From the graph it appears that the sum of the x coordinates of the 2 points of intersection are less than 0. I don't think any of the choices are correct.