Find two values of k, such that the line y = kx + 2 is tangent to the curve y = x^2 + 4x + 3

There will be intersection when x^2 + 4x + 3 = kx + 2. Our goal is to find the values of k which would only give one solution to this quadratic equation, which would make the lines 'tangent' to each other. First we rearrange the equation to get it in a familiar form: x^2 + (4-k)x + 1 = 0 To have one solution, the discriminant (b^2 - 4ac) must be zero. (4-k)^2 - 411 = 0 16 - 8k + k^2 - 4 = 0 k^2 - 8k +12 = 0 Factorising: (k-6)(k-2) = 0 So k = 6 and k = 2 are valid solutions to this problem.