Chapter 5 – Quadratic Functions Algebra 2

Table of Contents 5.5 - Complex Numbers and Roots 5.6 - The Quadratic Formula 5.7 – Solving Quadratic Inequalities 5.9 - Operations with Complex Numbers

5.5 - Complex Numbers and Roots Algebra 2

5.5 Algebra 2 (bell work) Pg. 350

5.5 Just Read You can see in the graph of f(x) = x2 + 1 below that f has no real zeros. If you solve the corresponding equation 0 = x2 + 1, you find that x = ,which has no real solutions.

Simplifying Square Roots of Negative Numbers 5.5 Example 1 Simplifying Square Roots of Negative Numbers Express the number in terms of i.

Express the number in terms of i 5.5

Math Joke Q: What’s a mathematician's favorite dessert? A: -1 ce cream 5.5 Math Joke Q: What’s a mathematician's favorite dessert? A: -1 ce cream

5.5 Example 2 Solving a Quadratic Equation with Imaginary Solutions Solve the equation 5x2 + 90 = 0

5.5 Solve the equation x2 + 48 = 0 9x2 + 25 = 0 x2 = –48 9x2 = –25

5.5 A complex number is a number that can be written in the form a + bi, where a and b are real numbers and i = . The set of real numbers is a subset of the set of complex numbers C. Every complex number has a real part a and an imaginary part b.

Equate the imaginary parts. Equate the real parts. 10 = –4y 4x = 2 5.5 Example 3 Equating Two Complex Numbers Find the values of x and y that make the equation true . Real parts 4x + 10i = 2 – (4y)i Imaginary parts Equate the imaginary parts. Equate the real parts. 10 = –4y 4x = 2 Solve for y. Solve for x.

Equate the imaginary parts. 2x = –8 –6 = 20y 5.5 Find the values of x and y that make each equation true 2x – 6i = –8 + (20y)i Real parts 2x – 6i = –8 + (20y)i Imaginary parts Equate the real parts. Equate the imaginary parts. 2x = –8 –6 = 20y x = –4 Solve for y. Solve for x.

f(x) = x2 + 10x + 26 g(x) = x2 + 4x + 12 5.5 Example 4 Finding Complex Zeros of Quadratic Functions Day 2 Find the zeros of the function. f(x) = x2 + 10x + 26 g(x) = x2 + 4x + 12 x2 + 10x + 26 = 0 x2 + 4x + 12 = 0 x2 + 10x + = –26 + x2 + 4x + = –12 + x2 + 10x + 25 = –26 + 25 x2 + 4x + 4 = –12 + 4 (x + 5)2 = –1 (x + 2)2 = –8

f(x) = x2 + 4x + 13 5.5 Find the zeros of the function. x = –2 ± 3i

Find each complex conjugate. 5.5 The solutions and are related. These solutions are a complex conjugate pair. Their real parts are equal and their imaginary parts are opposites. **The complex conjugate of any complex number a + bi is the complex number a – bi.** Example 5 Finding Complex Conjugates Find each complex conjugate. A. 8 + 5i B. 6i 8 + 5i 0 + 6i 0 – 6i 8 – 5i –6i

Find each complex conjugate 5.5 Find each complex conjugate A. 9 – i B. 9 + (–i) 9 – (–i) 9 + i C. –8i 0 + (–8)i 0 – (–8)i 8i

HW pg. 353 5.5- Day 1: 2-11, 18-25, 92-98 Day 2: 12-17, 29, 31, 36 Ch: 43, 58-65, 72, 75 HW Guidelines or ½ off

5.6 - The Quadratic Formula Algebra 2

5.6 Algebra 2 (bell work) You can use the Quadratic Formula to solve any quadratic equation that is written in standard form, including equations with real solutions or complex solutions. *Once you have your zeros, can always plug them back in and see if you get zero Videos

Don’t Copy 5.6

f(x)= 2x2 – 16x + 27 f(x) = x2 + 3x – 7 5.6 Example 1 Quadratic Functions with Real Zeros Find the zeros using the Quadratic Formula f(x)= 2x2 – 16x + 27 f(x) = x2 + 3x – 7 2x2 – 16x + 27 = 0 x2 + 3x – 7 = 0

5.6 Find the zeros of f(x)= x2 – 8x + 10 using the Quadratic Formula. Optional x2 – 8x + 10 = 0

5.6 Math Joke Q: What do you get when you cross an algebra class with prom? A: The Quadratic formal

f(x) = 4x2 + 3x + 2 g(x) = 3x2 – x + 8 5.6 Example 2 Quadratic Functions with Complex Zeros Find the zeros using the Quadratic Formula f(x) = 4x2 + 3x + 2 g(x) = 3x2 – x + 8 4x2 + 3x + 2 = 0

5.6 Day 2 The discriminant is part of the Quadratic Formula that you can use to determine the number of real roots of a quadratic equation.

5.6 Make sure the equation is in standard form before you evaluate the discriminant, b2 – 4ac. Caution!

5.6 Example 3 Analyzing Quadratic Equations by Using the Discriminant Find the type and number of solutions for the equation. x2 + 36 = 12x x2 + 40 = 12x x2 – 12x + 36 = 0 x2 – 12x + 40 = 0 b2 – 4ac b2 – 4ac (–12)2 – 4(1)(36) (–12)2 – 4(1)(40) 144 – 144 = 0 144 – 160 = –16 b2 – 4ac = 0 b2 –4ac < 0 The equation has one distinct real solution. The equation has two distinct nonreal complex solutions.

5.6 Find the type and number of solutions for the equation. x2 + 30 = 12x x2 – 12x + 30 = 0 b2 – 4ac (–12)2 – 4(1)(30) 144 – 120 = 24 b2 – 4ac > 0 The equation has two distinct real solutions.

5.6 Example 4 Application A pilot of a helicopter plans to release a bucket of water on a forest fire. The height y in feet of the water t seconds after its release is modeled by y = –16t2 – 2t + 500. The horizontal distance x in feet between the water and its point of release is modeled by x = 91t. The pilot’s altitude decreases, which changes the function describing the water’s height to y = –16t2 –2t + 400. To the nearest foot, at what horizontal distance from the target should the pilot begin releasing the water?

5.6 Step 1 Use the equation to determine how long it will take the water to hit the ground. Set the height of the water equal to 0 feet, and then use the quadratic formula for t. y = –16t2 – 2t + 400 0 = –16t2 – 2t + 400 t ≈ –5.063 or t ≈ 4.937 The time cannot be negative, so the water lands on a target about 4.937 seconds after it is released.

5.6 Step 2 The horizontal distance x in feet between the water and its point of release is modeled by x = 91t. Find the horizontal distance that the water will have traveled in this time. x = 91t x ≈ 91(4.937) x ≈ 449.267 x ≈ 449 The water will have traveled a horizontal distance of about 449 feet. Therefore, the pilot should start releasing the water when the horizontal distance between the helicopter and the fire is 449 feet.

5.6 Optional An athlete on a track team throws a shot put. The height y of the shot put in feet t seconds after it is thrown is modeled by y = –16t2 + 24.6t + 6.5. The horizontal distance x in between the athlete and the shot put is modeled by x = 29.3t. To the nearest foot, how far does the shot put land from the athlete?

5.6 Step 1 Use the first equation to determine how long it will take the shot put to hit the ground. Set the height of the shot put equal to 0 feet, and the use the quadratic formula to solve for t. y = –16t2 + 24.6t + 6.5 0 = –16t2 + 24.6t + 6.5 The time cannot be negative, so the shot put hits the ground about 1.8 seconds after it is released.

Step 2 Find the horizontal distance that the shot put will have traveled in this time. x = 29.3t x ≈ 29.3(1.77) x ≈ 51.86 x ≈ 52 The shot put will have traveled a horizontal distance of about 52 feet.

Properties of Solving Quadratic Equations 5.6 Properties of Solving Quadratic Equations Pg. 360

HW pg. 361 5.6- Day 1: 3-13 (Odd), 45-53 Day 2: 14-17, 54, 59, 76-78 Ch: 36, 37, 60 HW Guidelines or ½ off

5.7 - Solving Quadratic Inequalities Algebra 2

5.7 Algebra 2 (Bell work) Copy the steps below Pg. 366

5.7 Example 1 Graphing Quadratic Inequalities in Two Variables Graph y ≥ x2 – 7x + 10

Graph y ≥ 2x2 – 5x – 2 5.7

Graph y < –3x2 – 6x – 7 5.7

5.7 Math Joke Q: How can a fisherman determine how many fish he needs to catch to make a profit? A: By using a cod-ratic inequality

5.7 Example 2 Solving Quadratic Inequalities by Using Tables and Graphs Solve the inequality by using tables or graphs. x2 + 8x + 20 ≥ 5 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to x2 + 8x + 20 and Y2 equal to 5. Identify the values of x for which Y1 ≥ Y2.

5.7 The parabola is at or above the line when x is less than or equal to –5 or greater than or equal to –3. So, the solution set is x ≤ –5 or x ≥ –3 or (–∞, –5] U [–3, ∞). The table supports your answer. The number line shows the solution set. –6 –4 –2 0 2 4 6

5.7 Solve the inequality by using tables and graph. x2 + 8x + 20 < 5 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to x2 + 8x + 20 and Y2 equal to 5. Identify the values of which Y1 < Y2.

5.7 The parabola is below the line when x is greater than –5 and less than –3. So, the solution set is –5 < x < –3 or (–5, –3). The table supports your answer. The number line shows the solution set. –6 –4 –2 0 2 4 6

2x2 – 5x + 1 ≥ 1 5.7 Solve the inequality by using tables and graph. Optional 2x2 – 5x + 1 ≥ 1 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to 2x2 – 5x + 1 and Y2 equal to 1. Identify the values of which Y1 ≥ Y2.

5.7 The parabola is at or above the line when x is less than or equal to 0 or greater than or greater than or equal to 2.5. So, the solution set is (–∞, 0] U [2.5, ∞) or x ≤ 0 or x ≥ 2.5 The number line shows the solution set. –6 –4 –2 0 2 4 6

Solve the inequality x2 – 10x + 18 ≤ –3 by using algebra. 5.7 Example 3 Solving Quadratic Inequalities by Using Algebra Day 2 Solve the inequality x2 – 10x + 18 ≤ –3 by using algebra. Step 1 Write the related equation. x2 – 10x + 18 = –3 Step 2 Solve the equation for x to find the critical values. x2 –10x + 21 = 0 (x – 3)(x – 7) = 0 Step 3 Test an x-value in each interval. x – 3 = 0 or x – 7 = 0 x = 3 or x = 7

x2 – 10x + 18 ≤ –3 5.7 Step 3 Test an x-value in each interval. –3 –2 –1 0 1 2 3 4 5 6 7 8 9 Critical values Test points x2 – 10x + 18 ≤ –3 (2)2 – 10(2) + 18 ≤ –3 Try x = 2. x (4)2 – 10(4) + 18 ≤ –3  Try x = 4. (8)2 – 10(8) + 18 ≤ –3 x Try x = 8.

x2 – 6x + 10 ≥ 2 5.7 Solve the inequality by using algebra. Step 1 Write the related equation. x2 – 6x + 10 = 2 x2 – 6x + 8 = 0 Step 2 Solve the equation for x to find the critical values. (x – 2)(x – 4) = 0 x – 2 = 0 or x – 4 = 0 x = 2 or x = 4

x2 – 6x + 10 ≥ 2 5.7 Step 3 Test an x-value in each interval. –3 –2 –1 0 1 2 3 4 5 6 7 8 9 Critical values Test points x2 – 6x + 10 ≥ 2 (1)2 – 6(1) + 10 ≥ 2  Try x = 1. (3)2 – 6(3) + 10 ≥ 2 x Try x = 3. (5)2 – 6(5) + 10 ≥ 2  Try x = 5.

–2x2 + 3x + 7 < 2 5.7 Solve the inequality by using algebra. Optional –2x2 + 3x + 7 < 2 Step 1 Write the related equation. –2x2 + 3x + 7 = 2 Step 2 Solve the equation for x to find the critical values. –2x2 + 3x + 5 = 0 (–2x + 5)(x + 1) = 0 –2x + 5 = 0 or x + 1 = 0 x = 2.5 or x = –1

–2x2 + 3x + 7 < 2 5.7 Step 3 Test an x-value in each interval. –3 –2 –1 0 1 2 3 4 5 6 7 8 9 Critical values Test points –2x2 + 3x + 7 < 2 Try x = –2. –2(–2)2 + 3(–2) + 7 < 2  –2(1)2 + 3(1) + 7 < 2 x Try x = 1. –2(3)2 + 3(3) + 7 < 2  Try x = 3.

5.7 Example 4 Application Skip 15_16 The monthly profit P of a small business that sells bicycle helmets can be modeled by the function P(x) = –8x2 + 600x – 4200 Where x is the average selling price of a helmet. What range of selling prices will generate a monthly profit of at least $6000? List the important information: The profit must be at least $6000. The function for the business’s profit is P(x) = –8x2 + 600x – 4200. Write the inequality. –8x2 + 600x – 4200 ≥ 6000 Find the critical values by solving the related equation. –8x2 + 600x – 4200 = 6000 –8x2 + 600x – 10,200 = 0 –8(x2 – 75x + 1275) = 0

5.7 x ≈ 26.04 or x ≈ 48.96 –8(25)2 + 600(25) – 4200 ≥ 6000 Try x = 25. x 5800 ≥ 6000 –8(45)2 + 600(45) – 4200 ≥ 6000 Try x = 45. 6600 ≥ 6000  –8(50)2 + 600(50) – 4200 ≥ 6000 Try x = 50. 5800 ≥ 6000 x Write the solution as an inequality. The solution is approximately 26.04 ≤ x ≤ 48.96.

HW pg.370 5.7- Day 1: 2-4 (By Hand), 5-7 (Calc), 48-50, 72-74 Ch: 11, 14, 47, 51-53

5.9 - Operations With Complex Numbers Algebra 2

Just Read Algebra 2 (bell work) 5.9 Algebra 2 (bell work) The complex plane is a set of coordinate axes in which where x = real numbers, y = imaginary numbers Just Read The real axis corresponds to the x-axis, and the imaginary axis corresponds to the y-axis. Think of a + bi as x + yi. Helpful Hint

• • • • A. 2 – 3i B. –1 + 4i C. 4 + i D. –i 5.9 Example 1 Graphing Complex Numbers Graph each complex number. A. 2 – 3i • –1+ 4i B. –1 + 4i • 4 + i C. 4 + i • –i D. –i • 2 – 3i

5.9

A. |3 + 5i| B. |–13| C. |–7i| 5.9 Example 2 Determining the Absolute Value of Complex Numbers Find each absolute value. A. |3 + 5i| B. |–13| C. |–7i| |–13 + 0i| |0 +(–7)i| 7 13

Math Joke Q: Why did the imaginary number turn red? 5.9 Math Joke Q: Why did the imaginary number turn red? A: It ran out of i-drops

(4 + 2i) + (–6 – 7i) (5 –2i) – (–2 –3i) (1 – 3i) + (–1 + 3i) 5.9 Example 3 Adding and Subtracting Complex Numbers Add or subtract. Write the result in the form a + bi. (4 + 2i) + (–6 – 7i) (5 –2i) – (–2 –3i) (1 – 3i) + (–1 + 3i) (4 – 6) + (2i – 7i) (5 – 2i) + 2 + 3i (1 – 1) + (–3i + 3i) (5 + 2) + (–2i + 3i) –2 – 5i 7 + i

–2i(2 – 4i) (3 + 6i)(4 – i) 5.9 Example 5 Multiplying Complex Numbers Simplify by using the fact i2 = –1. Multiply. Write the result in the form a + bi. –2i(2 – 4i) (3 + 6i)(4 – i) –4i + 8i2 12 + 24i – 3i – 6i2 –4i + 8(–1) 12 + 21i – 6(–1) –8 – 4i 18 + 21i

5.9 Multiply. Write the result in the form a + bi. (2 + 9i)(2 – 9i) (–5i)(6i) 4 – 18i + 18i – 81i2 –30i2 4 – 81(–1) –30(–1) 85 30

Simplify –6i14 Simplify i63 5.9 Example 6 Evaluating Powers of i Day 2 Example 6 Evaluating Powers of i Simplify –6i14 Simplify i63 i63 = i  i62 –6i14 = –6(i2)7 = –6(–1)7 = i  (i2)31 = i  (–1)31 = i  –1 = –i = –6(–1) = 6

5.9 Simplify i42 Optional Simplify i42 = ( i2)21 = (–1)21 = –1

Dividing Complex Numbers 5.9 Example 7 Dividing Complex Numbers Simplify -Recall that expressions in simplest form cannot have square roots in the denominator (Lesson 1-3). -Because the imaginary unit represents a square root, you must rationalize any denominator that contains an imaginary unit. -To do this, multiply the numerator and denominator by the complex conjugate of the denominator.

Simplify Simplify 5.9

HW pg.386 5.9- Day 1: 2-5 (One Graph), 9-17 (Odd), 21-25 (Odd), 55-59 (Odd) Day 2: 27-35 (Odd), 85-99 (Odd), 121, 123 Ch: 103-112