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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
 #1
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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

 

 

cool cool cool

CPhill  Nov 7, 2017

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