O-Level Equations and inequalities: worked solution

Question

For y = 8x² + mx + m − 6 (m constant), find the values of m making the x-axis a tangent.

Saying that the x-axis is a tangent to the curve means the curve touches the line y = 0 at exactly one point. Setting y = 0 gives 8x² + mx + (m − 6) = 0, and this equation must have exactly one solution, that is a repeated root. Step 1: state the condition for a repeated root. A quadratic ax² + bx + c = 0 has two distinct roots when b² − 4ac > 0, no real roots when b² − 4ac < 0, and exactly one repeated root when b² − 4ac = 0. Step 2: identify a, b and c. Comparing 8x² + mx + (m − 6) with ax² + bx + c: a = 8, b = m, c = m − 6. Step 3: form the equation in m. b² − 4ac = 0 m² − 4 × 8 × (m − 6) = 0 m² − 32(m − 6) = 0 Expand the bracket, remembering that −32 multiplies both terms inside it: m² − 32m + 192 = 0. Step 4: solve this quadratic in m. Look for two numbers with product 192 and sum −32. Both must be negative; −8 and −24 work, since (−8) × (−24) = 192 and (−8) + (−24) = −32. m² − 32m + 192 = (m − 8)(m − 24) = 0. A product is zero when one factor is zero: m − 8 = 0 gives m = 8 m − 24 = 0 gives m = 24. So m = 8 or m = 24.