A line has equation \(y = 3x - 2k\) and a curve has equation \(y = x^2 - kx + 2\), where \(k\) is a constant.
Show that the line and the curve meet for all values of \(k\).
Solution
To find the points of intersection, set the equations equal: \(3x - 2k = x^2 - kx + 2\).
Rearrange to form a quadratic equation: \(x^2 - kx - 3x + 2 + 2k = 0\).
Simplify to: \(x^2 - (k+3)x + (2 + 2k) = 0\).
Use the discriminant \(b^2 - 4ac\) to determine if there are real solutions: \(b = -(k+3), a = 1, c = 2 + 2k\).
Calculate the discriminant: \((k+3)^2 - 4(1)(2 + 2k)\).
Simplify: \((k+3)^2 - 8 - 8k = (k-1)^2\).
Since \((k-1)^2 \geq 0\) for all \(k\), the discriminant is non-negative, ensuring real solutions.
Hence, the line and the curve meet for all values of \(k\).
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