Higher Quadratic Past Paper Questions

Higher Quadratic Past Paper Questions 2001 Paper I Q2 For what values of k does the equation x2 – 5x + (k + 6) = 0 have equal roots? a = 1 1x2 – 5x + (k + 6) = 0 b = – 5 c = (k + 6) Equal roots  b2 – 4ac = 0 (– 5)2 – 4(1)(k + 6) = 0 25 – 4k – 24 = 0 1 – 4k = 0 1 = 4k k = ¼ 3

Higher Quadratic Past Paper Questions Show that the equation (1 – 2k)x2 – 5kx – 2k = 0 has real roots for all integer values of k? a = 1 – 2k (1 – 2k)x2 – 5kx – 2k = 0 b = – 5k c = - 2k Real roots  b2 – 4ac ≥ 0 (– 5k)2 – 4(1 – 2k)(– 2k) ≥ 0 25k2 + 8k(1 – 2k) ≥ 0 25k2 + 8k – 16k2 ≥ 0 9k2 + 8k ≥ 0 For all integers 9k2 ≥ 0 & 9k2 ≥ 8k  always real roots for all integer values of k. 5

Higher Quadratic Past Paper Questions Show that the line with equation y = 2x + 1 does not intersect the parabola with equation y = x2 + 3x + 4 x2 + 3x + 4 = 2x + 1 a = 1 x2 + x + 3 = 0 b = 1 c = 3 Find value of b2 – 4ac (1)2 – 4(1)(3) = 1 – 12 = -11 < 0 As b2 – 4ac < 0  No intersection occurs. 5

Higher Quadratic Past Paper Questions Show that x = -1 is a solution of the cubic x3 + px2 + px + 1 = 0 1 Hence find the range of values of p for which all the roots are real 7 x3 + px2 + px + 1 = 0 1 p p 1 -1 As Remainder, R = 0  x = -1 is a solution & x3 + px2 + px + 1 = (x + 1)(1x2 + (p – 1)x + 1) -1 1–p -1 1 P–1 1 1

Higher Quadratic Past Paper Questions (b) Hence find the range of values of p for which all the roots are real 7 From (a) x3 + px2 + px + 1 = (x + 1)(1x2 + (p – 1)x + 1) a = 1 If real then b2 – 4ac ≥ 0 b = (p – 1) c = 1 (p – 1)2 – 4(1)(1) ≥ 0 p2 – 2p + 1 – 4 ≥ 0 p2 – 2p – 3 ≥ 0 (p – 3)(p + 1) ≥ 0  Real when p ≤ -1 & p ≥ 3 y -1 3 x 7

Higher Quadratic Past Paper Questions Express 2x2 + 4x – 3 in the form f(x) = a(x + b)2 + c 3 Write down the coordinaates of the turning point 1 2x2 + 4x – 3 = 2[x2 + 2x – 3/2] = 2[(x2 + 2x + 1) – 1 – 3/2] = 2[(x2 + 2x + 1) – 2/2 – 3/2] = 2[(x + 1)2 – 5/2] = 2(x + 1)2 – 5 Happy as + x2  (- 1, - 5) is Min Tpt 4

Higher Quadratic Past Paper Questions Find the range of values of k such that the equation kx2 – x – 1 = 0 has no real roots a = k kx2 – 1x – 1= 0 b = -1 c = -1 b2 – 4ac < 0  No Real Roots (-1)2 – 4(k)(-1) < 0 1 + 4k < 0 4k < – 1 k < – ¼ 4

Higher Quadratic Past Paper Questions 2008 Paper 1 Q10 Which of the following are true about the equation x2 + x + 1 = 0 (1) The roots are equal (2) The roots are real A = Neither B = (1) Only C = (2) Only D = Both are True a = 1 1x2 + 1x + 1= 0 b = 1 c = 1 Find nature of b2 – 4ac (1)2 – 4(1)(1) = 1 – 4 = -3 < 0 As b2 – 4ac < 0  Non real roots are not equal or real  A – Neither (1) or (2) are true 2

Higher Quadratic Past Paper Questions The graph has an equation of the form y = k(x – a)(x – b). What is the equation of the graph y 12 1 4 x Use roots at x = 1 & x = 4 for factors y = k(x – 1)(x – 4) 12 = k(0 – 1)(0 – 4) 12 = k(-1)(-4) 4k = 12 k = 3 y = 3(x – 1)(x – 4) A 2

Higher Quadratic Past Paper Questions 2x2 + 4x + 7 is expressed in the form 2(x + p)2 + q What is the value of q? 2 2x2 + 4x + 7 = 2[x2 + 2x + 7/2] = 2[(x2 + 2x + 1) – 1 + 7/2] = 2[(x2 + 2x + 1) – 2/2 + 7/2] = 2[(x + 1)2 + 5/2] = 2(x + 1)2 + 5 = 2(x + p)2 + q  q = 5 = A 2

Higher Quadratic Past Paper Questions Write x2 – 10x + 27 in the form (x + b)2 + c 2 Hence show that g(x) = 1/3 x3 – 5x2 + 27x – 2 is always increasing 4 (a) x2 – 10x + 27 = (x2 – 10x + 25) – 25 + 27 = (x – 5)2 + 2 Increasing  Differentiate g(x) = 1/3 x3 – 5x2 + 27x – 2 (b) g’(x) = x2 – 10x + 27 ** Use answer to (a) = (x – 5)2 + 2 (x – 5)2 + 2 > 0 for all x  function is always increasing 6

Higher Quadratic Past Paper Questions Total = 41 Marks