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A curve has equation y=x^2 + (3k - 4)x + 13 and a line has equation y = 2x + k, where k is constant. Show that the x-coordinate of any point of intersection of the line and curve satisfies the equation: x^2 + 3(k - 2)x + 13 - k = 0

When we deal with points of interception, this immediately indicates that these two equations have to equal. Therefore, begin by equaling these two equations: x^2 + (3k - 4)x + 13 = 2x + k Bring all figures to one side, like the answer shows you to do, and open out any brackets, so we can later simplify: x^2 + 3kx - 4x + 13 - k - 2x = 0 Simplify: x^2 + 3kx - 6x - k + 13 = 0 The answer shows that you now need to simplify the x terms, hence resulting in the final equation: x^2 + 3(k - 2)x + 13 - k = 0

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A curve has equation y=x^2 + (3k - 4)x + 13 and a line has equation y = 2x + k, where k is constant. Show that the x-coordinate of any point of intersection of the line and curve satisfies the equation: x^2 + 3(k - 2)x + 13 - k = 0

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