1. Write 15x2 + 6x = 14x2 – 12 in standard form. Lesson 4.8, For use with pages 292-299 1. Write 15x2 + 6x = 14x2 – 12 in standard form. ANSWER x2 + 6x +12 = 0 2. Evaluate b2 – 4ac when a = 3, b = –6, and c = 5. ANSWER –24

3. A student is solving an equation by completing the 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.” + _ ANSWER (x – 3)2 = 25

Questions 4.7

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

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. 

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

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

Write original equation. 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. x = – 4+ 42– 4(– 1)(– 5) 2(– 1) a = –1, b = 4, c = –5 x = – 4+ – 4 – 2 Simplify. – 4+ 2i x = – 2 Rewrite using the imaginary unit i. x = 2 + i Simplify. The solution is 2 + i and 2 – i. ANSWER

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 + i is shown. –(2 + i)2 + 4(2 + i) = 5 ? –3 – 4i + 8 + 4i = 5 ? 5 = 5 

Write original equation. GUIDED PRACTICE for Examples 1, 2, and 3 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. x = – b + b2 – 4ac 2a Quadratic formula x = – (– 6) + (– 6)2 – 4(1)(4) 2(1) a = 1, b = – 6, c = 4 x = + 3 + 20 2 Simplify.

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

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

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

Write original equation. GUIDED PRACTICE for Examples 1, 2, and 3 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. x = – b + b2 – 4ac 2a Quadratic formula x = – (5) + (5)2 – 4(– 5)(–7) 2(– 5) a = – 5, b = 5, c = – 7 –115 x = – 5 + – 10 Simplify.

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

EXAMPLE 4 Solve a quadratic inequality using a table 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 x are between –3 and 2, inclusive. The solution of the inequality is –3 ≤ x ≤ 2. ANSWER

EXAMPLE 5 Solve a quadratic inequality by graphing 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 x = – 1+ 12– 4(2)(– 4) 2(2) x = – 1+ 33 4 x 1.19 or x –1.69

EXAMPLE 5 Solve a quadratic inequality by graphing 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.

GUIDED PRACTICE for Examples 4 and 5 Solve the inequality 2x2 + 2x ≤ 3 using a table and using a graph. SOLUTION Rewrite the inequality as 2x2 + 2x – 3 ≤ 0. Then make a table of values. x -3 -2 -1.8 -1.5 -1 0.5 0.8 0.9 22 + 2x – 3 9 1 -0.1 0.42 The solution of the inequality is –1.8 ≤ x ≤ 0.82. ANSWER

EXAMPLE 6 Use a quadratic inequality as a model Robotics The number T of 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?

EXAMPLE 6 Use a quadratic inequality as a model SOLUTION You want to find the values of x for 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. There were more than 100 teams participating in the years 1997–2001. ANSWER

Write equation that corresponds to original inequality. 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 or x = 5 Zero product property

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: (– 4)2 –2(– 4) = 24 >15  12 –2(1) 5 –1 >15 62 –2(6) = 24 >15  The solution is x < – 3 or x > 5. ANSWER

GUIDED PRACTICE for Examples 6 and 7 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 x for which: T(x) > 200 7.51x2 – 16.4x + 35.0 > 200 7.51x2 – 16.4x – 165 > 0

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

Write equation that corresponds to original inequality. GUIDED PRACTICE for Examples 6 and 7 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 or x = 4 Zero product property

– 3 – 4 – 2 – 1 1 2 3 4 5 6 7 – 5 – 6 – 7 GUIDED PRACTICE for Examples 6 and 7 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. – 3 – 4 – 2 – 1 1 2 3 4 5 6 7 – 5 – 6 – 7 Test x = – 3: Test x = 2: Test x = 5: 2 (– 3)2 – 7 (– 3) > 4  2 (2)2 – 7 (2) > 4 2 (5)2 – 7 (3) > 4  The solution is x < – 0.5 or x > 4. ANSWER