O-Level Quadratic functions: worked solution

Question

Find the values of the constant m for which y = (m − 3)x² + mx + m stays entirely above the x-axis.

"Entirely above the x-axis" means y > 0 for every value of x: the curve never touches and never crosses y = 0. Two separate conditions must both hold. Condition 1: the parabola must open upwards. If the x² coefficient were negative the arms would go down to −∞ and the curve would drop below the axis. So m − 3 > 0 m > 3 (Note m − 3 cannot be 0 either, or the expression would not be a quadratic.) Condition 2: the curve must not meet the x-axis at all, so the equation (m − 3)x² + mx + m = 0 must have no real roots. That happens when the discriminant is negative: b² − 4ac < 0, with a = m − 3, b = m and c = m m² − 4(m − 3)(m) < 0 Expand the product: 4(m − 3)(m) = 4(m² − 3m) = 4m² − 12m m² − (4m² − 12m) < 0 m² − 4m² + 12m < 0 −3m² + 12m < 0 Multiply every term by −1. Multiplying an inequality by a negative number reverses the inequality sign: 3m² − 12m > 0 Factorise the left-hand side: 3m(m − 4) > 0 The critical values are m = 0 and m = 4. Since the coefficient of m² is positive, the expression is above zero outside the two critical values: m < 0 or m > 4 Now combine the two conditions, because both must be true at the same time. Condition 1 requires m > 3, so the branch m < 0 is impossible and is rejected. The part that satisfies both m > 3 and m > 4 is m > 4. So the curve lies entirely above the x-axis when m > 4.