O-Level Equations and inequalities: worked solution

Question

Find the values of p for which the parabola y = x² − x − 2 stays completely above the line y = p(x + 2).

"Completely above" means the curve's y value beats the line's y value for every single x, with no crossing and no touching: x² - x - 2 > p(x + 2) for all values of x Bring everything to the left-hand side: x² - x - 2 - p(x + 2) > 0 x² - x - 2 - px - 2p > 0 Group the x terms and the constant terms: x² - (1 + p)x - (2 + 2p) > 0 for all x This is a quadratic in x whose x² coefficient is 1, which is positive, so its graph is a U-shape. A U-shaped curve lies entirely above the x-axis exactly when it never reaches the axis, i.e. when it has no real roots. The condition for no real roots is discriminant < 0, that is b² - 4ac < 0 Here a = 1, b = -(1 + p) and c = -(2 + 2p), so [-(1 + p)]² - 4(1)[-(2 + 2p)] < 0 (1 + p)² + 4(2 + 2p) < 0 Expand: 1 + 2p + p² + 8 + 8p < 0 Collect like terms: p² + 10p + 9 < 0 Factorise: two numbers that multiply to 9 and add to 10 are 1 and 9: (p + 1)(p + 9) < 0 As a quadratic in p this is again a U-shape, so it is negative between its two roots. The roots are p = -1 and p = -9, so -9 < p < -1 The inequalities are strict because the discriminant must be strictly negative; if it were zero the line would touch the curve rather than stay below it.