O-Level Equations and inequalities: worked solution

Question

Find the values of k for which kx² + (k+1)x + k is negative for all x. [3]

'Negative for all x' means the whole parabola lies below the x-axis, which needs two things to be true at once. Condition 1: the curve opens downwards, so the coefficient of x² must be negative: k < 0 Condition 2: the curve never touches or crosses the x-axis, so the equation kx² + (k+1)x + k = 0 has no real roots, which means the discriminant is negative. Here a = k, b = k + 1 and c = k: (k + 1)² − 4(k)(k) < 0 Expand the bracket: k² + 2k + 1 − 4k² < 0 −3k² + 2k + 1 < 0 Multiply every term by −1, and REVERSE the inequality sign because it is a multiplication by a negative: 3k² − 2k − 1 > 0 Factorise: 3k² − 2k − 1 = (3k + 1)(k − 1) (3k + 1)(k − 1) > 0 The critical values are k = −1/3 and k = 1. This is a U-shaped expression in k, and a U shape is positive OUTSIDE its roots: k < −1/3 or k > 1 Combine the two conditions. Both must hold: k < 0 AND (k < −1/3 or k > 1). The branch k > 1 fails k < 0, so it is rejected; the branch k < −1/3 already satisfies k < 0. k < −1/3