Chapter Closure
Sign In
Substitute values
| x=2± 4i/2 | |
|---|---|
| x_1=2+4i/2 | x_2=2-4i/2 |
| x_1=1+2i | x_2=1-2i |
Now, consider Equation (I). y=2x We can substitute x=1+2i and x=1-2i into the above equation to find the values for y.
| x | 2x | y=2x |
|---|---|---|
| 1+2i | 2( 1+2i) | 2+4i |
| 1-2i | 2( 1-2i) | 2-4i |
Therefore, the solutions to the system are (1+2i, 2+4i) and (1-2i, 2-4i).
y=2x y=x^2+5 Note that Equation (I) represents the equation of the line and Equation (II) — the equation of the quadratic function. The solutions to the system are the points where the graphs of the line and the parabola intersect. Solutions (1+2i, 2+4i), (1-2i, 2-4i) We can see that the coordinates of the points are complex numbers. Since there are no real-valued points of intersection, the graphs of the line and the parabola do not intersect on a coordinate plane.
