A linear-nonlinear system pairs a line with a curve, so it may have more than one solution. Graph both equations in Desmos to see every intersection, then set their expressions for equal. If the question asks for the sum of all possible -values, the resulting quadratic’s coefficients give an exact answer without relying on rounded graph coordinates.
Hints
- Hint 1
An intersection is a point on both graphs, so its coordinates satisfy both equations. A line can cross a parabola twice. Click both meeting points and look at their first coordinates, which are the possible -values.
- Hint 2
At each intersection, the two equations give the same . Set their right sides equal, then move the terms into the form .
- Hint 3
The roots of a quadratic are the -values that make it equal . Their sum is , not . What are and after you combine the two equations?
Step-by-step
Graph the system, then use the root sum
Step 1Find every intersection
Type both equations into Desmos and click both places where the graphs meet. The points are near and . An intersection lies on both graphs, so both first coordinates are possible values of . The displayed coordinates are rounded; use algebra for the exact sum.
- Step 2
Make one equation in
At either intersection, both equations give the same , so set their right sides equal:
- Step 3
Collect the -terms
Subtract from both sides:
- Step 4
Put the quadratic in standard form
Add to both sides:
The term is the one you’ll use to find the sum.
- Step 5
Find the exact sum
A root is an -value that makes the quadratic equal . If its roots are and , expanding gives . So in , the middle coefficient satisfies , which means the roots add to . Here and :
That includes all possible values of , not only one intersection. Grid in 1.