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# For what negative value of $k$ is there exactly one solution to the system of equations

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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*}

michaelcai  Nov 7, 2017
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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

CPhill  Nov 7, 2017

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