kx + y = 1
y + -x^2 + k
In the system of equation above, k is a constant. When the equations are graphed in the xy - plane, the graph intersect at exactly two points. Which of the following CANNOT be the value of k?
A. 3
B. 2
C. 1
D. 0
+33616 Alternatively, you could do it algebraically. Equate the two expressions for y:
x2+k = 1-kx
x2 + kx + k-1 = 0. (1)
The discriminant of this quadratic is k2-4*1*(k-1) or k2-4k+4. This is zero when k = 2.
Since there is only a single solution to equation (1) when the discriminant is zero, k = 2 doesn’t allow the original two equations to intersect in more than one point.
+33616 Alternatively, you could do it algebraically. Equate the two expressions for y:
x2+k = 1-kx
x2 + kx + k-1 = 0. (1)
The discriminant of this quadratic is k2-4*1*(k-1) or k2-4k+4. This is zero when k = 2.
Since there is only a single solution to equation (1) when the discriminant is zero, k = 2 doesn’t allow the original two equations to intersect in more than one point.
The graph above might help (even though it is a little busy!).
