O-Level Quadratic functions: worked solution

Question

Find the range of the constant p for which y = px² - 4x + p - 3 is positive for all real x.

"y = px² − 4x + p − 3 is positive for all real x" means the graph lies entirely above the x-axis, never touching it. Two conditions are needed for that. Condition 1: the parabola must open upwards (a U shape), otherwise it would eventually go below the axis. So the coefficient of x² must be positive: p > 0. Condition 2: the curve must never touch or cross the x-axis, so the quadratic has no real roots, which means the discriminant is negative. With a = p, b = −4 and c = p − 3: b² − 4ac < 0 (−4)² − 4(p)(p − 3) < 0 16 − 4p(p − 3) < 0. Expand the bracket: 16 − 4p² + 12p < 0. Divide every term by −4. Dividing an inequality by a negative number reverses the inequality sign: −4 + p² − 3p > 0 p² − 3p − 4 > 0. Factorise the left-hand side. Two numbers that multiply to −4 and add to −3 are −4 and +1: (p − 4)(p + 1) > 0. The critical values are p = 4 and p = −1. For a U-shaped expression the product is positive outside the roots, so p < −1 or p > 4. Finally combine this with Condition 1, p > 0. The branch p < −1 contains no values that are also greater than 0, so it is rejected. The branch p > 4 already satisfies p > 0, so it survives. Therefore p > 4.