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### Transcript

• 1. Section 1.2Quadratic Equations Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 2. Quadratic Equations Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 3. Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 4. Solve the equation: x &#x2212; 4x = 0 2 x( x &#x2212; 4) = 0 x = 0 or x + 4 = 0 x = 0 or x &#x2212; 4 The solution set is { 0, 4} Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 5. Solve the equation: x2 = 6 &#x2212; x x + x&#x2212;6 = 0 2 ( x + 3)( x &#x2212; 2) = 0 x = &#x2212;3 or x = 2 The solution set is { &#x2212;3, 2} Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 6. Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 7. Solve each equation. ( x + 3) 2 (a) x = 7 2 (b) =9 x+3= &#xB1; 9 x=&#xB1; 7 x + 3 = &#xB1;3 x = 7 or x = &#x2212; 7 x + 3 = 3 or x + 3 = &#x2212;3 x = 0 or x = &#x2212;6 {The solution set is &#x2212; 7, 7 } The solution set is { 0, &#x2212;6} Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 8. Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 9. Solve by completing the square: 2x2 + 6x &#x2212; 5 = 0 2x + 6x = 5 2 3 19 x+ =&#xB1; 2 4 2x2 6x 5 + = 2 2 2 19 3 x=&#xB1; &#x2212; 9 5 9 2 2x + 3x + = + 2 4 2 4 19 &#x2212; 3 19 &#x2212; 3 x= or &#x2212; 2 2 2&#xF8EB; 3 &#xF8F6; 19&#xF8EC;x+ &#xF8F7; =&#xF8ED; 2&#xF8F8; 4 Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 10. Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 11. Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 12. Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 13. Use the quadratic formula to find the real solutions if any, of the equation 2x &#x2212; 4x +1 = 0 2 ax + bx + c = 0 2 b 2 &#x2212; 4ac = (&#x2212;4) 2 &#x2212; 4(2)(1) = 16 &#x2212; 8 = 8 b &#x2212; 4ac &gt; 0 so there are two real solutions 2 which can be found using the quadratic formula. Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 14. Use the quadratic formula to find the real solutions if any, of the equation 2x &#x2212; 4x +1 = 0 2 &#x2212;b &#xB1; b &#x2212; 4ac 2 x= 2a &#x2212;(&#x2212;4) &#xB1; 8 4&#xB1; 8 2&#xB1; 2 x= = = 2(2) 4 2 &#xF8F1;2 + 2 2 &#x2212; 2 &#xF8FC; &#xF8F4; &#xF8F4; The solution set is &#xF8F2; , &#xF8FD; &#xF8F4; 2 &#xF8F3; 2 &#xF8F4;&#xF8FE; Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 15. Use the quadratic formula to find the real solutions if any, of the equation 1 2 x &#x2212; 6 x + 18 = 0 x 2 &#x2212; 12 x + 36 = 0 2 ax 2 + bx + c = 0 b 2 &#x2212; 4ac = (&#x2212;12) 2 &#x2212; 4(1)(36) = 144 &#x2212; 144 = 0 b &#x2212; 4ac = 0 so there is a repeated solution 2 which can be found using the quadratic formula. Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 16. Use the quadratic formula to find the real solutions if any, of the equation 1 2 x &#x2212; 6 x + 18 = 0 x &#x2212; 12 x + 36 = 0 2 2 ax 2 + bx + c = 0 &#x2212;b &#xB1; b 2 &#x2212; 4acx= 2a &#x2212;(&#x2212;12) &#xB1; 0 12 x= = =6 The solution set is { 6} 2(1) 2 Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 17. Use the quadratic formula to find the real solutions if any, of the equation 2x + 3 = 2x 2 Since b &#x2212; 4ac &lt; 0, 2 2x &#x2212; 2x + 3 = 0 2 there is no real solution. ax 2 + bx + c = 0 b &#x2212; 4ac = ( &#x2212;2 ) &#x2212; 4(2)(3) 2 2 = 4 &#x2212; 24 = &#x2212;20 Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 18. Use the quadratic formula to find the real solutions if any, of the equation 1 2 &#x2212;1 &#xB1; 49 &#x2212;1 &#xB1; 7 6+ &#x2212; 2 = 0 x= = x x 2(6) 12 6 x2 + x &#x2212; 2 = 0 &#x2212;1 + 7 1 = = 12 2 b &#x2212; 4ac = ( 1) &#x2212; 4(6)(&#x2212;2) 2 2 &#x2212;1 &#x2212; 7 2 = =&#x2212; = 1 + 48 = 49 12 3Since b 2 &#x2212; 4ac &gt; 0, &#xF8F1;1 2&#xF8FC; The solution set is &#xF8F2; , &#x2212; &#xF8FD;there are two real solutions. &#xF8F3;2 3&#xF8FE; Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 19. Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 20. Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 21. Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.
• 22. 9 ( x &#x2212; 18 ) = 144 ( x &#x2212; 18) x &#x2212; 18 = &#xB1;4 2 2 = 16 22 centimeters by 22 centimetersx = 18 &#xB1; 4 = 22 or 14 Copyright &#xA9; 2012 Pearson Education, Inc. Publishing as Prentice Hall.