A line has equation `y = kx + 6` and a curve has equation

`y = x^2 + 3x + 2k` , where k is a constant.

Find the two values of k for which the line is a tangent to the curve.

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When line is tangent to the curve the gradient of tangent is equal to the gradient of line.

At tangent points;

`kx+6 = x^2+3x+2k`

`0 = x^2 +(3-k)x+(2k-6)` -------------(1)

At tangent points gradient is given by the first derivative of curve.

`dy/dx = 2x+3`

Since gradient of line = gradient of tangent;

`2x+3 = k rarr x= (k-3)/2`

From (1)

`0 = ((k-3)/2)^2 +((3-k)(k-3))/2 +2k-6`

`0 = (k^2-6k+9)/4 - (k^2-6k+9)/2 +2k-6`

`0 = k^2-6k+9-(2k^2-12k+18)+8k-24`

`0=-k^2+6k-9+8k-24`

`0=k^2-14k+33`

`0=k^2-11k-3k+33`

`0=k(k-11)-3(k-11)`

`0=(k-11)(k-3)`

*So the required k values are k=11 and k=3*

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