O-Level Equations and inequalities

From 4y = 2x + 1, x = (4y - 1)/2. Substitute into 3y - x = 4xy: 3y - (4y-1)/2 = 4y(4y-1)/2. Left side = (6y - 4y + 1)/2 = (2y+1)/2; right side = 2y(4y-1) = 8y² - 2y. So (2y+1)/2 = 8y² - 2y, i.e. 2y + 1 = 16y² - 4y, giving 16y² - 6y - 1 = 0. Factorising/quadratic formula: y = (6 +/- sqrt(36 + 64))/32 = (6 +/- 10)/32, so y = 1/2 or y = -1/8. When y = 1/2, x = (2 - 1)/2 = 1/2. When y = -1/8, x = (-1/2 - 1)/2 = -3/4. Hence A = (1/2, 1/2) and B = (-3/4, -1/8).

For 2x² - cx + (c² + 6) = 0, the discriminant is D = (-c)² - 4(2)(c² + 6) = c² - 8c² - 48 = -7c² - 48. Since c² >= 0 for every real c, we have -7c² <= 0, so D = -7c² - 48 <= -48 < 0. Because D < 0 for all real c, the equation has no real roots for every real value of c.

A line and a curve meet where they have the same y for the same x, so set the two expressions for y equal: kx² + 3 = −5x − 2 Collect every term on the left to get a quadratic equation in x: kx² + 5x + 3 + 2 = 0 kx² + 5x + 5 = 0 The number of points of intersection is the number of real roots of this quadratic, and that is decided by the discriminant b² − 4ac: b² − 4ac > 0 means two distinct real roots, so two intersection points, b² − 4ac = 0 means one repeated root, so the line is a tangent, b² − 4ac < 0 means no real roots, so no intersection. Here a = k, b = 5 and c = 5, so b² − 4ac = 5² − 4 × k × 5 = 25 − 20k Now bring in the condition k < 1. Multiplying an inequality by the negative number −20 reverses it: k < 1 −20k > −20 Add 25 to both sides: 25 − 20k > 25 − 20 25 − 20k > 5 Since 5 > 0, the discriminant is greater than 5 and therefore certainly positive for every value of k with k < 1. A positive discriminant gives two distinct real roots, so yes: for every k < 1 the line y = −5x − 2 cuts the curve y = kx² + 3 at two distinct points.