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### Transcript

• 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 + &#xF04B; = &#x2212;c + &#xF04B; b &#x2212;c 2 x + x +&#xF04B;= +&#xF04B; a a 2 b &#x2212;c &#xF8EB; b &#xF8F6; &#xF8EB; b &#xF8F6; x + x+&#xF8EC; &#xF8F7; = +&#xF8EC; &#xF8F7; a a &#xF8ED; 2a &#xF8F8; &#xF8ED; 2a &#xF8F8; 2 2
• 4. The Quadratic Formula Solve for x by completing the square. ax 2 + bx + c = 0 ax 2 + bx + &#xF04B; = &#x2212;c + &#xF04B; b &#x2212;c x2 + x + &#xF04B; = +&#xF04B; a a 2 b &#x2212;c &#xF8EB; b &#xF8F6; &#xF8EB; b &#xF8F6; x2 + x + &#xF8EC; &#xF8F7; = +&#xF8EC; &#xF8F7; a a &#xF8ED; 2a &#xF8F8; &#xF8ED; 2a &#xF8F8; b b2 b 2 &#x2212; 4ac x2 + x + 2 = a 4a 4a 2 2 2 b &#xF8F6; b 2 &#x2212; 4ac &#xF8EB; &#xF8EC;x+ &#xF8F7; = 2a &#xF8F8; 4a 2 &#xF8ED;
• 5. The Quadratic Formula Solve for x by completing the square. b b2 b 2 &#x2212; 4ac x + x+ 2 = a 4a 4a 2 2 2 b &#xF8F6; b 2 &#x2212; 4ac &#xF8EB; &#xF8EC;x+ &#xF8F7; = 2a &#xF8F8; 4a 2 &#xF8ED; b b 2 &#x2212; 4ac x+ =&#xB1; 2a 4a 2 &#x2212;b b 2 &#x2212; 4ac x= &#xB1; 2a 2a &#x2212; b &#xB1; b 2 &#x2212; 4ac x= 2a
• 6. The Quadratic Formula Solve for x by completing the square. &#x2212; b &#xB1; b &#x2212; 4ac x= 2a 2
• 7. How does it work Equation: 3x 2 + 5 x + 1 = 0 a=3 b=5 c =1 &#x2212; b &#xB1; b 2 &#x2212; 4ac x= 2a
• 8. How does it work Equation: 3x + 5 x + 1 = 0 a=3 b=5 c =1 2 &#x2212; b &#xB1; b 2 &#x2212; 4ac x= 2a x= &#x2212; ( 5) &#xB1; ( 5) 2 &#x2212; 4( 3)(1) 2( 3) &#x2212; 5 &#xB1; 25 &#x2212; 12 x= 6 x= &#x2212; 5 &#xB1; 13 &#x2212; 5 13 = &#xB1; 6 6 6
• 9. The Discriminant The number in the square root of the quadratic formula. b &#x2212; 4ac 2 Given x &#x2212; 5 x + 6 = 0 2 ( &#x2212; 5) &#x2212; 4(1)( 6 ) 25 &#x2212; 24 = 1 2
• 10. The Discriminant b &#x2212; 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 &#xB1; 5 ).
• 11. The Discriminant b 2 &#x2212; 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&#x2019;s put all of that b information in a chart. Value of Discriminant Type and Number of Roots D &gt; 0, D is a perfect square 2 real, rational roots (ex: x= 2 and x= -4) D &gt; 0, D NOT a perfect square 2 real, Irrational roots (x = &#x221A;13 x= -&#x221A;13) D=0 1 real, rational root (double root) (ex: x = 5) D&lt;0 2 complex roots (complex conjugates) (x = 2 &#xB1; 3i ) 2 &#x2212; 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 &#x2212; 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 &#x2212; 4 = 0 2
• 19. Solve using the Quadratic formula x &#x2212; 8 x = 33 2
• 20. Solve using the Quadratic formula x &#x2212; 8 x = 33 2 x &#x2212; 8 x &#x2212; 33 = 0 2 x= &#x2212; ( &#x2212; 8) &#xB1; ( &#x2212; 8) &#x2212; 4(1)( &#x2212; 33) 2(1) 2
• 21. Solve using the Quadratic formula x 2 &#x2212; 8 x = 33 x 2 &#x2212; 8 x &#x2212; 33 = 0 x= &#x2212; ( &#x2212; 8) &#xB1; ( &#x2212; 8) 2 &#x2212; 4(1)( &#x2212; 33) 2(1) 8 &#xB1; 196 8 &#xB1; 14 x= = 2 2 8 + 14 22 x= = = 11 2 2 8 &#x2212; 14 &#x2212; 6 x= = = &#x2212;3 2 2
• 22. Solve using the Quadratic formula x &#x2212; 34 x + 289 = 0 2
• 23. Solve using the Quadratic formula x &#x2212; 34 x + 289 = 0 2 x= &#x2212; ( &#x2212; 34 ) &#xB1; ( &#x2212; 34) 2(1) 2 &#x2212; 4(1)( 289 )
• 24. Solve using the Quadratic formula x &#x2212; 34 x + 289 = 0 2 x= &#x2212; ( &#x2212; 34) &#xB1; ( &#x2212; 34) 2 &#x2212; 4(1)( 289) 2(1) 34 &#xB1; 1156 &#x2212; 1156 x= 2 34 &#xB1; 0 34 x= = = 17 2 2
• 25. Solve using the Quadratic formula x &#x2212; 6x + 2 = 0 2
• 26. Solve using the Quadratic formula x &#x2212; 6x + 2 = 0 2 x= &#x2212; ( &#x2212; 6) &#xB1; ( &#x2212; 6) 2(1) 2 &#x2212; 4(1)( 2 ) 6 &#xB1; 36 &#x2212; 8 6 &#xB1; 28 x= = 2 2 6 2 7 x= &#xB1; = 3&#xB1; 7 2 2
• 27. Solve using the Quadratic formula x 2 + 13 = 6 x x 2 &#x2212; 6 x + 13 = 0 x= &#x2212; ( &#x2212; 6) &#xB1; ( &#x2212; 6) 2 &#x2212; 4(1)(13) 2(1)
• 28. Solve using the Quadratic formula x 2 + 13 = 6 x x 2 &#x2212; 6 x + 13 = 0 x= &#x2212; ( &#x2212; 6) &#xB1; ( &#x2212; 6) 2 &#x2212; 4(1)(13) 2(1) 6 &#xB1; 36 &#x2212; 52 6 &#xB1; &#x2212; 16 x= = 2 2
• 29. Solve using the Quadratic formula x 2 + 13 = 6 x x 2 &#x2212; 6 x + 13 = 0 x= &#x2212; ( &#x2212; 6) &#xB1; ( &#x2212; 6) 2 &#x2212; 4(1)(13) 2(1) 6 &#xB1; 36 &#x2212; 52 6 &#xB1; &#x2212; 16 x= = 2 2 6 &#xB1; 4i 6 4 x= = &#xB1; i 2 2 2 x = 3 &#xB1; 2i