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Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
Quadratic eq and discriminant
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Quadratic eq and discriminant

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  • 1. 5-6 The Quadratic Formula and the Discriminant p.292-297
  • 2. We have a number of different way of finding the roots if a quadratic equations #1. Making a table #2. Factoring #3. Completing the Square Now a new way that comes from completing the square. The Quadratic Formula
  • 3. The Quadratic Formula Solve for x by completing the square. ax 2 + bx + c = 0 ax 2 + bx +  = −c +  b −c 2 x + x += + a a 2 b −c  b   b  x + x+  = +  a a  2a   2a  2 2
  • 4. The Quadratic Formula Solve for x by completing the square. ax 2 + bx + c = 0 ax 2 + bx +  = −c +  b −c x2 + x +  = + a a 2 b −c  b   b  x2 + x +   = +  a a  2a   2a  b b2 b 2 − 4ac x2 + x + 2 = a 4a 4a 2 2 2 b  b 2 − 4ac  x+  = 2a  4a 2 
  • 5. The Quadratic Formula Solve for x by completing the square. b b2 b 2 − 4ac x + x+ 2 = a 4a 4a 2 2 2 b  b 2 − 4ac  x+  = 2a  4a 2  b b 2 − 4ac x+ =± 2a 4a 2 −b b 2 − 4ac x= ± 2a 2a − b ± b 2 − 4ac x= 2a
  • 6. The Quadratic Formula Solve for x by completing the square. − b ± b − 4ac x= 2a 2
  • 7. How does it work Equation: 3x 2 + 5 x + 1 = 0 a=3 b=5 c =1 − b ± b 2 − 4ac x= 2a
  • 8. How does it work Equation: 3x + 5 x + 1 = 0 a=3 b=5 c =1 2 − b ± b 2 − 4ac x= 2a x= − ( 5) ± ( 5) 2 − 4( 3)(1) 2( 3) − 5 ± 25 − 12 x= 6 x= − 5 ± 13 − 5 13 = ± 6 6 6
  • 9. The Discriminant The number in the square root of the quadratic formula. b − 4ac 2 Given x − 5 x + 6 = 0 2 ( − 5) − 4(1)( 6 ) 25 − 24 = 1 2
  • 10. The Discriminant b − 4ac 2 The Discriminant can be negative, positive or zero If the Discriminant is positive then there are: 2 real answers. If the square root is not a prefect square ( for example 25 ), then there will be 2 irrational roots ( for example 2 ± 5 ).
  • 11. The Discriminant b 2 − 4ac The Discriminant can be negative, positive or zero If the Discriminant is positive, there are 2 real answers. If the Discriminant is zero, there is 1 real answer. If the Discriminant is negative, there are 2 complex answers. complex answer have i.
  • 12. Let’s put all of that b information in a chart. Value of Discriminant Type and Number of Roots D > 0, D is a perfect square 2 real, rational roots (ex: x= 2 and x= -4) D > 0, D NOT a perfect square 2 real, Irrational roots (x = √13 x= -√13) D=0 1 real, rational root (double root) (ex: x = 5) D<0 2 complex roots (complex conjugates) (x = 2 ± 3i ) 2 − 4ac Sample Graph of Related Function
  • 13. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0
  • 14. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root
  • 15. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2 -11, Two complex roots x + 8x − 4 = 0 2 80, Two irrational roots
  • 16. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2
  • 17. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2 -11, Two complex roots
  • 18. Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2 -11, Two complex roots x + 8x − 4 = 0 2
  • 19. Solve using the Quadratic formula x − 8 x = 33 2
  • 20. Solve using the Quadratic formula x − 8 x = 33 2 x − 8 x − 33 = 0 2 x= − ( − 8) ± ( − 8) − 4(1)( − 33) 2(1) 2
  • 21. Solve using the Quadratic formula x 2 − 8 x = 33 x 2 − 8 x − 33 = 0 x= − ( − 8) ± ( − 8) 2 − 4(1)( − 33) 2(1) 8 ± 196 8 ± 14 x= = 2 2 8 + 14 22 x= = = 11 2 2 8 − 14 − 6 x= = = −3 2 2
  • 22. Solve using the Quadratic formula x − 34 x + 289 = 0 2
  • 23. Solve using the Quadratic formula x − 34 x + 289 = 0 2 x= − ( − 34 ) ± ( − 34) 2(1) 2 − 4(1)( 289 )
  • 24. Solve using the Quadratic formula x − 34 x + 289 = 0 2 x= − ( − 34) ± ( − 34) 2 − 4(1)( 289) 2(1) 34 ± 1156 − 1156 x= 2 34 ± 0 34 x= = = 17 2 2
  • 25. Solve using the Quadratic formula x − 6x + 2 = 0 2
  • 26. Solve using the Quadratic formula x − 6x + 2 = 0 2 x= − ( − 6) ± ( − 6) 2(1) 2 − 4(1)( 2 ) 6 ± 36 − 8 6 ± 28 x= = 2 2 6 2 7 x= ± = 3± 7 2 2
  • 27. Solve using the Quadratic formula x 2 + 13 = 6 x x 2 − 6 x + 13 = 0 x= − ( − 6) ± ( − 6) 2 − 4(1)(13) 2(1)
  • 28. Solve using the Quadratic formula x 2 + 13 = 6 x x 2 − 6 x + 13 = 0 x= − ( − 6) ± ( − 6) 2 − 4(1)(13) 2(1) 6 ± 36 − 52 6 ± − 16 x= = 2 2
  • 29. Solve using the Quadratic formula x 2 + 13 = 6 x x 2 − 6 x + 13 = 0 x= − ( − 6) ± ( − 6) 2 − 4(1)(13) 2(1) 6 ± 36 − 52 6 ± − 16 x= = 2 2 6 ± 4i 6 4 x= = ± i 2 2 2 x = 3 ± 2i

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