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# X2 t01 02 solving quadratics (2013)

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### X2 t01 02 solving quadratics (2013)

1. 1. Solving Quadratics e.g . x 2  3 x  7  0
2. 2. Solving Quadratics e.g . x 2  3 x  7  0  3  9  28 2  3   19  2 x
3. 3. Solving Quadratics e.g . x 2  3 x  7  0  3  9  28 2  3   19  2  3  19i  2  3  19i  3  19i x or x  2 2 x
4. 4. Solving Quadratics e.g . x 2  3 x  7  0  3  9  28 x 2  3   19  2  3  19i  2  3  19i  3  19i x or x  2 2 OR by completing the square x 2  3x  7  0
5. 5. Solving Quadratics e.g . x 2  3 x  7  0  3  9  28 x 2  3   19  2  3  19i  2  3  19i  3  19i x or x  2 2 OR by completing the square x 2  3x  7  0 2 3  19 x    0 2 4  2  x  3   19 i 2  0   2 4 
6. 6. Solving Quadratics e.g . x 2  3 x  7  0  3  9  28 x 2  3   19  2  3  19i  2  3  19i  3  19i x or x  2 2 OR by completing the square x 2  3x  7  0 2 3  19 x    0 2 4  2  x  3   19 i 2  0   2 4  3 19  3 19   i  x   i  0 x  2 2  2 2  
7. 7. Solving Quadratics e.g . x 2  3 x  7  0  3  9  28 x 2  3   19  2  3  19i  2  3  19i  3  19i x or x  2 2 OR by completing the square x 2  3x  7  0 2 3  19 x    0 2 4  2  x  3   19 i 2  0   2 4  3 19  3 19   i  x   i  0 x  2 2  2 2   3 19 3 19 x  i or x    i 2 2 2 2
8. 8. Creating Quadratics All quadratics with real coefficients have conjugate pairs as solutions
9. 9. Creating Quadratics All quadratics with real coefficients have conjugate pairs as solutions  x  z   x  z   x2  ( z  z) x  z z
10. 10. Creating Quadratics All quadratics with real coefficients have conjugate pairs as solutions  x  z   x  z   x2  ( z  z) x  z z  x 2  2 Re( z ) x  z z
11. 11. Creating Quadratics All quadratics with real coefficients have conjugate pairs as solutions  x  z   x  z   x2  ( z  z) x  z z  x 2  2 Re( z ) x  z z iii  Find the quadratic equation with roots 4  i and 4-i
12. 12. Creating Quadratics All quadratics with real coefficients have conjugate pairs as solutions  x  z   x  z   x2  ( z  z) x  z z  x 2  2 Re( z ) x  z z iii  Find the quadratic equation with roots 4  i and 4-i   8   17
13. 13. Creating Quadratics All quadratics with real coefficients have conjugate pairs as solutions  x  z   x  z   x2  ( z  z) x  z z  x 2  2 Re( z ) x  z z iii  Find the quadratic equation with roots 4  i and 4-i   8   17  equation is x 2  8 x  17  0
14. 14. Finding Square Roots a  ib  x  iy
15. 15. Finding Square Roots a  ib  x  iy  a  ib   x  iy 2 a 2  2abi  b 2  x  iy
16. 16. Finding Square Roots a  ib  x  iy If x  iy  a  ib  a  ib   x  iy then a 2  b 2  x a 2  2abi  b 2  x  iy 2ab  y 2
17. 17. Finding Square Roots a  ib  x  iy If x  iy  a  ib  a  ib   x  iy then a 2  b 2  x a 2  2abi  b 2  x  iy 2ab  y 2 e.g. Find  12  16i
18. 18. Finding Square Roots a  ib  x  iy If x  iy  a  ib  a  ib   x  iy then a 2  b 2  x a 2  2abi  b 2  x  iy 2ab  y 2 e.g. Find  12  16i a 2  b 2  12 2ab  16 8 b a
19. 19. Finding Square Roots a  ib  x  iy If x  iy  a  ib  a  ib   x  iy then a 2  b 2  x a 2  2abi  b 2  x  iy 2ab  y 2 e.g. Find  12  16i a 2  b 2  12 64 2 a  2  12 a 2ab  16 8 b a
20. 20. Finding Square Roots a  ib  x  iy If x  iy  a  ib  a  ib   x  iy then a 2  b 2  x a 2  2abi  b 2  x  iy 2ab  y 2 e.g. Find  12  16i a 2  b 2  12 64 2 a  2  12 a a 4  12a 2  64  0 2ab  16 8 b a
21. 21. Finding Square Roots a  ib  x  iy If x  iy  a  ib  a  ib   x  iy then a 2  b 2  x a 2  2abi  b 2  x  iy 2ab  y 2 e.g. Find  12  16i a 2  b 2  12 64 2 a  2  12 a a 4  12a 2  64  0 a 2  4 a 2  16   0 a 2  4 or a 2  16 2ab  16 8 b a
22. 22. Finding Square Roots a  ib  x  iy If x  iy  a  ib  a  ib   x  iy then a 2  b 2  x a 2  2abi  b 2  x  iy 2ab  y 2 e.g. Find  12  16i a 2  b 2  12 64 2 a  2  12 a a 4  12a 2  64  0 a 2  4 a 2  16   0 a 2  4 or a 2  16 a  2 no real solutions  b  4 2ab  16 8 b a
23. 23. Finding Square Roots a  ib  x  iy If x  iy  a  ib  a  ib   x  iy then a 2  b 2  x a 2  2abi  b 2  x  iy 2ab  y 2 e.g. Find  12  16i a 2  b 2  12 64 2 a  2  12 a a 4  12a 2  64  0 a 2 2ab  16 8 b a  4 a 2  16   0 a 2  4 or a 2  16 a  2 no real solutions  b  4  12  16i  2  4i 
24. 24. OR a  ib  x  iy
25. 25. OR a  ib  x  iy a 2  b2  x 2ab  y
26. 26. OR a  ib  x  iy a b  x 2ab  y 2 x  y  a  b 2 2  2 2  2 2  4a 2b 2  a 4  2a 2 b 2  b 4  4a 2b 2 = a 4  2a 2 b 2  b 4
27. 27. OR a  ib  x  iy a b  x 2ab  y 2 x  y  a  b 2 2 If x  iy  a  ib then a 2  b 2  x a 2  b2  x2  y 2  2 2  2 2  4a 2b 2  a 4  2a 2 b 2  b 4  4a 2b 2 = a 4  2a 2 b 2  b 4
28. 28. OR a  ib  x  iy a b  x 2ab  y 2 x  y  a  b 2 2 If x  iy  a  ib then a 2  b 2  x a 2  b2  x2  y 2  2 2  2 2  4a 2b 2  a 4  2a 2 b 2  b 4  4a 2b 2 = a 4  2a 2 b 2  b 4 Note: y > 0, then a and b have same sign y < 0, then a and b have different sign
29. 29. OR a  ib  x  iy a b  x 2ab  y 2 x  y  a  b 2 2 If x  iy  a  ib then a 2  b 2  x a 2  b2  x2  y 2 e.g. Find  12  16i  2 2  2 2  4a 2b 2  a 4  2a 2 b 2  b 4  4a 2b 2 = a 4  2a 2 b 2  b 4 Note: y > 0, then a and b have same sign y < 0, then a and b have different sign
30. 30. OR a  ib  x  iy a b  x 2ab  y 2 x  y  a  b 2 2 If x  iy  a  ib then a 2  b 2  x a 2  b2  x2  y 2 e.g. Find  12  16i a 2  b 2  12 a 2  b 2  20  2 2  2 2  4a 2b 2  a 4  2a 2 b 2  b 4  4a 2b 2 = a 4  2a 2 b 2  b 4 Note: y > 0, then a and b have same sign y < 0, then a and b have different sign
31. 31. OR a  ib  x  iy a b  x 2ab  y 2 x  y  a  b 2 2 If x  iy  a  ib then a 2  b 2  x a 2  b2  x2  y 2 e.g. Find  12  16i a 2  b 2  12 a 2  b 2  20 2a 2 8 a2  4 a2  2 2  2 2  4a 2b 2  a 4  2a 2 b 2  b 4  4a 2b 2 = a 4  2a 2 b 2  b 4 Note: y > 0, then a and b have same sign y < 0, then a and b have different sign
32. 32. OR a  ib  x  iy a b  x 2ab  y 2 x  y  a  b 2 2 If x  iy  a  ib then a 2  b 2  x a 2  b2  x2  y 2 e.g. Find  12  16i a 2  b 2  12 a 2  b 2  20 2a 2 8 a2  4 a  2 b  4  2 2  2 2  4a 2b 2  a 4  2a 2 b 2  b 4  4a 2b 2 = a 4  2a 2 b 2  b 4 Note: y > 0, then a and b have same sign y < 0, then a and b have different sign
33. 33. OR a  ib  x  iy a b  x 2ab  y 2 x  y  a  b 2 2 If x  iy  a  ib then a 2  b 2  x a 2  b2  x2  y 2  2 2  2 2  4a 2b 2  a 4  2a 2 b 2  b 4  4a 2b 2 = a 4  2a 2 b 2  b 4 Note: y > 0, then a and b have same sign y < 0, then a and b have different sign e.g. Find  12  16i a 2  b 2  12 a 2  b 2  20 8 2a 2 a2  4 a  2 b  4  12  16i  2  4i 
34. 34. (ii ) Find 5  4i
35. 35. (ii ) Find 5  4i a 2  b2  5 a 2  b 2  41
36. 36. (ii ) Find 5  4i a 2  b2  5 a 2  b 2  41 2a 2  5  41 1 2 a  5  41 2  a  5  41 2 1 a 10  2 41 2
37. 37. (ii ) Find 5  4i a 2  b2  5 a 2  b 2  41 2a 2  5  41 1 2 a  5  41 2  a  5  41 2 1 a 10  2 41 2    1  b  5  41  5 2 1  5  41 2 2  5  41 b 2 1 10  2 41 b 2
38. 38. (ii ) Find 5  4i a 2  b2  5 a 2  b 2  41 2a 2  5  41 1 2 a  5  41 2  a  5  41 2 1 a 10  2 41 2 1 5  4i   2  10  2    1  b  5  41  5 2 1  5  41 2 2  5  41 b 2 1 10  2 41 b 2 41  i 10  2 41 
39. 39. (iii ) Solve z 2  (2  i ) z  (2  2i )  0
40. 40. (iii ) Solve z 2  (2  i ) z  (2  2i )  0 z   2  i   2  i   4  2  2i  2 2   2  i   4  4i  1  8  8i 2   2  i   5  12i 2
41. 41. (iii ) Solve z 2  (2  i ) z  (2  2i )  0 z   2  i   2  i   4  2  2i  2 2   2  i   4  4i  1  8  8i 2   2  i   5  12i 2 5  12i a 2  b 2  5 a 2  b 2  13 2a 2 8 a2  4 a  2 b  3
42. 42. (iii ) Solve z 2  (2  i ) z  (2  2i )  0 z    2  i   2  i   4  2  2i  2 2   2  i   4  4i  1  8  8i 2   2  i   5  12i 2   2  i    2  3i  2 5  12i a 2  b 2  5 a 2  b 2  13 2a 2 8 a2  4 a  2 b  3
43. 43. (iii ) Solve z 2  (2  i ) z  (2  2i )  0 z    2  i   2  i   4  2  2i  2 2   2  i   4  4i  1  8  8i 2   2  i   5  12i 2   2  i    2  3i  2 2i 4  4i  or 2 2  i or  2  2i 5  12i a 2  b 2  5 a 2  b 2  13 2a 2 8 a2  4 a  2 b  3
44. 44. Cambridge: Exercise 1B; 1 to 4 ace etc, 5, 6, 7 cf, 9, 10, 16 ac