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For what negative value of \(k\) is there exactly one solution to the system of equations \(y=2x^2+kx+6, y=-x+4\)?

 Sep 6, 2020
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We need the two functions to be equal, and their slopes to be equal.

 

\(2x^2+kx+6=-x+4\\4x+k=-1\\ \text{so:}\\2x^2+(k+1)x+2=0\\4x+(k+1)=0\)

 

Use the second to replace k+1 in the first and solve for x

\(2x^2-4x^2+2=0\\2x^2 = 2\\x^2=1\\x_1=1\\x_2=-1\)

 

Now \(k=-1-4x\)  so I'll leave it to you to decide which value of x to substitute in here in order to obtain a negative value for k.

 Sep 6, 2020

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