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Lesson 4.8 , For use with pages 292-299

Lesson 4.8 , For use with pages 292-299. 1. Write 15 x 2 + 6 x = 14 x 2 – 12 in standard form. ANSWER. x 2 + 6 x +12 = 0. 2. Evaluate b 2 – 4 ac when a = 3, b = –6, and c = 5. –24. ANSWER. _. +. Lesson 4.8 , For use with pages 292-299.

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Lesson 4.8 , For use with pages 292-299

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  1. Lesson 4.8, For use with pages 292-299 1. Write15x2 + 6x = 14x2 – 12in standard form. ANSWER x2 + 6x +12 = 0 2. Evaluateb2 – 4acwhena = 3, b = –6, andc = 5. –24 ANSWER

  2. _ + Lesson 4.8, For use with pages 292-299 3. A student is solving an equation by completing the square. Write the step in the solution that appears just before “(x – 3) = 5.” (x – 3)2 = 25 ANSWER

  3. Questions 4.7

  4. x = 2a –b+b2– 4ac –3+32– 4(1)(–2) x = 2(1) 17 17 –3+ –3+ x = x = 2 2 ANSWER The solutions are 0.56and 17 –3– x = –3.56. 2 EXAMPLE 1 Solve an equation with two real solutions Solvex2 + 3x = 2. x2 + 3x = 2 Write original equation. x2+ 3x – 2 = 0 Write in standard form. Quadratic formula a = 1, b = 3, c = –2 Simplify.

  5. EXAMPLE 1 Solve an equation with two real solutions CHECK Graph y = x2 + 3x – 2 and note that the x-intercepts are about 0.56 and about –3.56. 

  6. 30+(–30)2 – 4(25)(9) x = 30+ 0 2(25) x = 50 3 x = 5 3 ANSWER 5 The solution is EXAMPLE 2 Solve an equation with one real solutions Solve25x2 – 18x = 12x – 9. 25x2 – 18x = 12x – 9. Write original equation. 25x2 – 30x + 9 = 0. Write in standard form. a = 25, b = –30, c = 9 Simplify. Simplify.

  7. EXAMPLE 2 Solve an equation with one real solutions CHECK Graph y = –5x2 – 30x + 9 and note that the only x-intercept is 0.6 = . 3  5

  8. – 4+42– 4(– 1)(– 5) x = – 4+– 4 2(– 1) x = – 2 – 4+ 2i x = – 2 ANSWER The solution is 2 + i and 2 – i. EXAMPLE 3 Solve an equation with imaginary solutions Solve–x2 + 4x = 5. –x2 + 4x = 5 Write original equation. –x2 + 4x – 5 = 0. Write in standard form. a = –1, b = 4, c = –5 Simplify. Rewrite using the imaginary unit i. x = 2 +i Simplify.

  9. ? –(2 + i)2 + 4(2 + i) = 5 ? –3 – 4i + 8 + 4i = 5 5 = 5  EXAMPLE 3 Solve an equation with imaginary solutions CHECK Graph y = 2x2 + 4x – 5. There are no x-intercepts. So, the original equation has no real solutions. The algebraic check for the imaginary solution 2 + iis shown.

  10. –b+b2– 4ac x = 2a – (–6)+(–6)2– 4(1)(4) x = 2(1) +3+ 20 x = 2 for Examples 1, 2, and 3 GUIDED PRACTICE Use the quadratic formula to solve the equation. x2= 6x – 4 SOLUTION x2= 6x – 4 Write original equation. x2–6x + 4 = 0 Write in standard form. Quadratic formula a = 1, b = – 6, c = 4 Simplify.

  11. 20 3+ x = The solutions are and = 3 + 5 2 20 3– x = = 3 – 5 2 for Examples 1, 2, and 3 GUIDED PRACTICE ANSWER

  12. –b+b2– 4ac x = 2a – (–12)+(–12)2– 4(4)(9) x = 2(4) 12+ 0 x = 8 for Examples 1, 2, and 3 GUIDED PRACTICE Use the quadratic formula to solve the equation. 4x2 – 10x = 2x – 9 SOLUTION Write original equation. 4x2 – 10x = 2x – 9 Write in standard form. 4x2 – 12x + 9= 0 Quadratic formula a = 4, b = – 12, c = 9 Simplify.

  13. 1 = The solution is 3 1 2 2 for Examples 1, 2, and 3 GUIDED PRACTICE ANSWER

  14. –b+b2– 4ac x = 2a – (5)+(5)2– 4(–5)(–7) x = 2(–5) –5+ –115 x = – 10 for Examples 1, 2, and 3 GUIDED PRACTICE Use the quadratic formula to solve the equation. 7x – 5x2 – 4 = 2x + 3 SOLUTION 7x – 5x2 – 4 = 2x + 3 Write original equation. – 5x2 + 5x – 7 = 0 Write in standard form. Quadratic formula a = – 5, b = 5, c = – 7 Simplify.

  15. –5+i 115 x = – 10 5+i 115 x = 10 ANSWER 5+ i 5–i 115 115 The solutions are and . 10 10 for Examples 1, 2, and 3 GUIDED PRACTICE Rewrite using the imaginary unit i. Simplify.

  16. ANSWER The solution of the inequality is –3 ≤ x ≤ 2. Solve a quadratic inequality using a table EXAMPLE 4 Solve x2 + x ≤ 6 using a table. SOLUTION Rewrite the inequality as x2 + x – 6 ≤ 0. Then make a table of values. Notice that x2 + x –6 ≤ 0 when the values of xare between –3 and 2, inclusive.

  17. – 1+ 12– 4(2)(– 4) x = – 1+33 2(2) x = 4 x 1.19 orx –1.69 Solve a quadratic inequality by graphing EXAMPLE 5 Solve 2x2 + x – 4 ≥ 0 by graphing. SOLUTION The solution consists of the x-values for which the graph of y = 2x2 + x – 4 lies on or above the x-axis. Find the graph’s x-intercepts by letting y = 0 and using the quadratic formula to solve for x. 0 = 2x2 + x – 4

  18. Solve a quadratic inequality by graphing EXAMPLE 5 Sketch a parabola that opens up and has 1.19 and –1.69 as x-intercepts. The graph lies on or above the x-axis to the left of (and including) x = – 1.69 and to the right of (and including) x = 1.19. ANSWER The solution of the inequality is approximately x ≤ – 1.69 or x ≥ 1.19.

  19. ANSWER The solution of the inequality is –1.8 ≤ x ≤ 0.82. for Examples 4 and 5 GUIDED PRACTICE Solve the inequality 2x2 + 2x ≤ 3using a table and using a graph. SOLUTION Rewrite the inequality as 2x2 + 2x –3≤ 0. Then make a table of values.

  20. Robotics EXAMPLE 6 Use a quadratic inequality as a model The number Tof teams that have participated in a robot-building competition for high school students can be modeled by T(x) = 7.51x2 –16.4x + 35.0, 0 ≤ x ≤ 9 Where x is the number of years since 1992. For what years was the number of teams greater than 100?

  21. ANSWER There were more than 100 teams participating in the years 1997–2001. EXAMPLE 6 Use a quadratic inequality as a model SOLUTION You want to find the values of xfor which: T(x) > 100 7.51x2 – 16.4x + 35.0 > 100 7.51x2 – 16.4x – 65 > 0 Graph y = 7.51x2 – 16.4x – 65 on the domain 0 ≤ x ≤ 9. The graph’s x-intercept is about 4.2. The graph lies above the x-axis when 4.2 < x ≤ 9.

  22. EXAMPLE 7 Solve a quadratic inequality algebraically Solve x2 – 2x > 15 algebraically. SOLUTION First, write and solve the equation obtained by replacing > with = . x2 – 2x = 15 Write equation that corresponds to original inequality. x2 – 2x – 15 = 0 Write in standard form. (x + 3)(x – 5) = 0 Factor. x = – 3 orx = 5 Zero product property

  23. (– 4)2–2(– 4) = 24 >15 62 –2(6) = 24 >15 12 –2(1) 5 –1 >15   ANSWER The solution is x < – 3 orx > 5. EXAMPLE 7 Solve a quadratic inequality algebraically The numbers – 3 and 5 are the critical x-values of the inequality x2 – 2x > 15. Plot – 3 and 5 on a number line, using open dots because the values do not satisfy the inequality. The critical x-values partition the number line into three intervals. Test an x-value in each interval to see if it satisfies the inequality. Test x = – 4: Test x = 1: Test x = 6:

  24. for Examples 6 and 7 GUIDED PRACTICE Robotics Use the information in Example 6 to determine in what years at least 200 teams participated in the robot-building competition. SOLUTION You want to find the values of xfor which: T(x) > 200 7.51x2 – 16.4x + 35.0 > 200 7.51x2 – 16.4x – 165 > 0

  25. ANSWER There were more than 200 teams participating in the years 1998 – 2001. for Examples 6 and 7 GUIDED PRACTICE Graph y = 7.51x2 – 16.4x – 165 on the domain 0 ≤ x ≤ 9.

  26. for Examples 6 and 7 GUIDED PRACTICE Solve the inequality 2x2 – 7x = 4 algebraically. SOLUTION First, write and solve the equation obtained by replacing > with 5. 2x2 – 7x = 4 Write equation that corresponds to original inequality. 2x2 – 7x –4= 0 Write in standard form. (2x + 1)(x – 4) = 0 Factor. x = – 0.5 orx = 4 Zero product property

  27. 5 – 2 0 1 2 3 4 6 – 6 – 7 – 3 – 1 – 5 7 – 4 2 (– 3)2 – 7 (– 3)> 4 2 (2)2 – 7 (2)> 4 2 (5)2 – 7 (3)> 4   ANSWER The solution is x < – 0.5 orx > 4. for Examples 6 and 7 GUIDED PRACTICE The numbers 4 and –0.5 are the critical x-values of the inequality 2x2 – 7x > 4 . Plot 4 and –0.5 on a number line, using open dots because the values do not satisfy the inequality. The critical x-values partition the number line into three intervals. Test an x-value in each interval to see if it satisfies the inequality. Test x = – 3: Test x = 2: Test x = 5:

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