After you set $x^2 + 2x - 1$ equal to $-x + 5$ and move all terms to one side, you should get a quadratic equation. Write it clearly in the form $ax^2 + bx + c = 0$ and identify $a$, $b$, and $c$.

Instead of solving the quadratic completely, recall: for $ax^2 + bx + c = 0$, how can you find the product of the two solutions using only $a$ and $c$?

Use your identified values of $a$ and $c$ from the quadratic you formed to compute $c/a$. That value is the product of the $x$-coordinates of the intersection points.

Algebra is faster for this question, but Desmos can verify the intersections. Enter $y=x^2+2x-1$ and $y=-x+5$ on separate expression lines.

Click each point where the parabola and line intersect. Desmos displays the two intersection coordinates and confirms that there are two corresponding $x$-values.

For an exact calculator check, enter

This multiplies the two solutions of $x^2+3x-6=0$ and verifies the algebraic product.

At intersection points, the $y$-values of the two graphs are equal, so set the right sides equal:

Move all terms to one side to get a quadratic equation:

The $x$-coordinates of the intersection points are the solutions of $x^2+3x-6=0$.

You are asked for the product of the two $x$-coordinates, not the coordinates themselves.

For a quadratic of the form

with solutions $r_1$ and $r_2$, the product of the solutions is

Therefore, the product can be found directly from the leading coefficient and constant term.

Compare $x^2+3x-6=0$ with $ax^2+bx+c=0$:

• $a=1$
• $b=3$
• $c=-6$

The product of the two roots, which are the $x$-coordinates of the intersection points, is

So, the product of the $x$-coordinates of the intersection points is $-6$.