+619 For what negative value of $k$ is there exactly one solution to the system of equations\(\begin{align*} y &= 2x^2 + kx + 6 \\ y &= -x + 4? \end{align*}\)
+130071 y = 2x^2 + kx + 6
y = 4 - x
Set these equal
4 - x = 2x^2 + kx + 6 rearrange as
2x^2 + (k + 1)x + 2 = 0
If there is one solution, the discriminant must = 0
So we have that
(k + 1)^2 - 4(2) (2) = 0
(k + 1) ^2 - 16 = 0
( k + 1)^2 = 16 take the negative root
k + 1 = -4
k = - 5
