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University of Southampton School of Education 24 week mathematics subject knowledge enhancement course Developing Subject Knowledge – Quadratics answers 1. Using factorisation, solve each equation : a) x² – 4x + 3 = 0 (x -1)(x- 3) = 0 x = 1 or 3 b) x(x – 1) = 42 x²- x -42 = 0 (x + 6)(x- 7) = 0 x = -6 or 7 d) 6 + 23x – 4x² = 0 4x² -23x- 6 = 0 (4x + 1)(x- 6) = 0 x = -¼ or 6 c) x² = 3x x²-3x = 0 x(x- 3) = 0 x = 0 or 3 2. By completing the square, find the maximum or minimum value of y and the value of x for which this occurs. State whether your value of y is a maximum or a minimum in each case. a) y = x – 2x + 7 y = (x -1)² -1 + 7 y = (x -1)² + 6 y = 6 at x = 1, minimum b) y = x² + 10x + 35 y = (x + 5)² -25 + 35 y = (x + 5)² + 10 y = 10 at x = -5, minimum c) y = – x² + 4x + 4 y =- [x² -4x] + 4 y =- [(x -2)² - 4] + 4 y = -(x -2)² + 8 y = 8 at x = 2, maximum d) y = 2x² + 8x + 5 y = 2x² + 8x + 5 y = 2(x² + 4x) + 5 y = 2((x + 2)² -4) + 5 y = 2(x + 2)² -8 + 5 y = 2(x + 2)² -3 y = -3 at x =- 2, minimum 3. Use the quadratic formula to solve each equation, giving your answers as simply as possible in terms of surds where appropriate. a) x² + 4x + 1 = 0 x = – 2 ±√3 b) – t² + 8t + 4 = 0 c) y² – 20y + 91 = 0 t = 4 ± 2√5 d) 5 – y – y² = 0 y = –½ (1 ±√21) y = 7 or 13 4. By evaluating the discriminant, determine whether the roots of each equation are real and distinct, real and equal or not real. a) x² + 2x – 7 = 0 b² – 4ac = 32 real and distinct b) x² + x + 3 = 0 c) x² + 14x + 49 = 0 b² – 4ac = – 11 not real b² – 4ac = 0 real and equal d) 5x² + 8x + 3 = 0 b² – 4ac = 4 real and distinct 5. Find the coordinates of the points of intersection of the given straight line and curve in each case. a) y = x + 2 y = x² – 4 (–2, 0) and (3, 5) b) y = 4x + 11 y = x² + 3x – 1 (–3, –1) and (4, 27) c) 2x – y = 5 x² + y² = 25 (0, –5) and (4, 3) 6. Giving your answers in terms of surds, solve the equation x4 – x² – 6 = 0 (x² + 2)(x² -3) = 0 x² = -2 [no solutions] or x² = 3 x = ±√3 7. Solve the equation x = 8√x – 15 let y = √x hence x = y² x = 8√x – 15 becomes y² = 8y – 15 y² – 8y + 15 = 0 (y – 5)(y – 3) = 0 y = 5 or y = 3 x = 25 or x = 9 8. The diagram on the right shows the curve with equation y = 2x² + px + q, where p and q are constants, which crosses the x-axis at the points with coordinates (–2, 0) and (3, 0). a) Show that p = –2 and find the value of q Method 1 y = a(x + 2)(x – 3) (equation of graph must be of this form to cross x-axis at –2 and 3 = a(x² – x – 6) = ax² – ax – 6a ie coefficients of x² and x must be same magnitude but of opposite sign. Since coefficient of x² is 2, coefficient of x (p) must also be –2 q = – 6a = –12 Method 2 When x = –2, y = 0 and when x = 3, y = 0 Substituting into : y = 2x² + px + q gives 0 = 8 – 2p + q and 0 = 18 +3p + q Solving these simultaneous equations gives p = –2 and q = –12 b) Find the coordinates of the vertex of the curve Method 1 y = 2x² – 2x – 12 = 2[x² – x] – 12 = 2[(x – ½)² – ¼]– 12 = 2(x – ½)² – ½ – 12 = 2(x – ½)² – 12½ Vertex at (½,–12½) Method 2 Equation of line of symmetry : x = ½ (3 + -2) = ½ y = 2x² – 2x – 12 = – 12½