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If: y = 4x^2+kx+1 and y = x-8

Then: 4x^2+kx+1 = x-8 => 4x^2+kx-x+1+8 = 0 => 4x^2+(kx-x)+9 = 0

For the equation to be tangent the discriminant of b^2-4ac must = 0

So: (k-1)^2-4(4*9) = 0 => (k-1)^2-144 = 0 => (k-1)^2 = 144

Square root both sides: k-1 = -or+12 => k = 1-or+12

Therefore possible values of k are: -11 or 13

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Q: What are the possible values of k when the line y equals x-8 is a tangent to the curve y equals4x squared plus kx plus 1?
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