Download presentation

Presentation is loading. Please wait.

1
**Simplify each expression.**

Warm Up Simplify each expression. 2. 1. 3. Find the zeros of each function. 4. f(x) = x2 – 18x + 16 5. f(x) = x2 + 8x – 24

2
Ch. 4.6 : I Can define and use imaginary and complex numbers and solve quadratic equations with complex roots Do Now: Find the zeros of each function. 1. f(x) = x2 – 18x + 16 2. f(x) = x2 + 8x – 24 Success Criteria: Today’s Agenda Check HW Notes Assignment I can use complex numbers I can apply the conjugates

3
Ch. 4.6 : I Can define and use imaginary and complex numbers and solve quadratic equations with complex roots Do Now: Find the zeros of each function. 2. x2 + 8x +20 = 0 1. 3x = 0 Success Criteria: Today’s Agenda Notes Assignment I can use complex numbers I can apply the conjugates

4
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. However, you can find solutions if you define the square root of negative numbers, which is why imaginary numbers were invented. The imaginary unit i is defined as You can use the imaginary unit to write the square root of any negative number.

6
**Example 1A: Simplifying Square Roots of Negative Numbers**

Express the number in terms of i. Factor out –1. Product Property. Simplify. Multiply. Express in terms of i.

7
**Example 1B: Simplifying Square Roots of Negative Numbers**

Express the number in terms of i. Factor out –1. Product Property. Simplify. Express in terms of i.

8
Check It Out! Example 1a Express the number in terms of i.

9
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.

10
**Example 3A: Adding and Subtracting Complex Numbers**

Add or subtract. Write the result in the form a + bi. (4 + 2i) + (–6 – 7i) (4 – 6) + (2i – 7i) Add real parts and imaginary parts. –2 – 5i

11
**Example 3B: Adding and Subtracting Complex Numbers**

Add or subtract. Write the result in the form a + bi. (5 –2i) – (–2 –3i) (5 – 2i) i Distribute. (5 + 2) + (–2i + 3i) Add real parts and imaginary parts. 7 + i

12
Check It Out! Example 3a Add or subtract. Write the result in the form a + bi. (–3 + 5i) + (–6i) (–3) + (5i – 6i) Add real parts and imaginary parts. –3 – i

13
Check It Out! Example 3b Add or subtract. Write the result in the form a + bi. 2i – (3 + 5i) (2i) – 3 – 5i Distribute. (–3) + (2i – 5i) Add real parts and imaginary parts. –3 – 3i

14
**Example 5A: Multiplying Complex Numbers**

Multiply. Write the result in the form a + bi. –2i(2 – 4i) –4i + 8i2 Distribute. –4i + 8(–1) Use i2 = –1. –8 – 4i Write in a + bi form.

15
**Example 5B: Multiplying Complex Numbers**

Multiply. Write the result in the form a + bi. (3 + 6i)(4 – i) i – 3i – 6i2 Multiply. i – 6(–1) Use i2 = –1. i Write in a + bi form.

16
**Example 5C: Multiplying Complex Numbers**

Multiply. Write the result in the form a + bi. (2 + 9i)(2 – 9i) 4 – 18i + 18i – 81i2 Multiply. 4 – 81(–1) Use i2 = –1. 85 Write in a + bi form.

17
**The solutions and are related**

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. If a quadratic equation with real coefficients has nonreal roots, those roots are complex conjugates. When given one complex root, you can always find the other by finding its conjugate. Helpful Hint

18
**Example 5: Finding Complex Zeros of Quadratic Functions**

Find each complex conjugate. B. 6i A i 0 + 6i 8 + 5i Write as a + bi. Write as a + bi. 0 – 6i Find a – bi. 8 – 5i Find a – bi. –6i Simplify. C. 9 – i D. 9 + (–i) Write as a + bi. Write as a + bi. 9 – (–i) Find a – bi. Find a – bi. 9 + i Simplify.

19
**Use complex conjugates to simplify**

Multiply by the conjugate. Distribute. Use i2 = –1. Simplify.

20
**Use complex conjugates to simplify**

Multiply by the conjugate. Distribute. Use i2 = –1. Simplify.

21
Lesson Quiz: Part II Perform the indicated operation. Write the result in the form a + bi. 1. (2 + 4i) + (–6 – 4i) –4 2. (5 – i) – (8 – 2i) –3 + i 3. (2 + 5i)(3 – 2i) 4. 3 + i i 5. Simplify i31. –i

22
Assignment #47 p #9-12, 18 – 26, 27, 29, 31

23
Ch. 4.6 : I Can define and use imaginary and complex numbers and solve quadratic equations with complex roots Do Now: The base of a triangle is 3 cm longer than its altitude. The area of the triangle is 35 cm2. Find the altitude. Hint: draw the triangle and use formula. Success Criteria: Today’s Agenda Notes Assignment I can use complex numbers I can apply the conjugates

24
**Example 2A: Solving a Quadratic Equation with Imaginary Solutions**

Solve the equation. Take square roots. Express in terms of i. Check x2 = –144 –144 (12i)2 144i 2 144(–1) x2 = –144 –144 144(–1) 144i 2 (–12i)2

25
**Example 2B: Solving a Quadratic Equation with Imaginary Solutions**

Solve the equation. 5x = 0 Add –90 to both sides. Divide both sides by 5. Take square roots. Express in terms of i. 5x = 0 5(18)i 2 +90 90(–1) +90 Check

26
Check It Out! Example 2a Solve the equation. x2 = –36 x = 0 x2 = –48 9x = 0 9x2 = –25

27
**Example 4A: Finding Complex Zeros of Quadratic Functions**

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 = – x2 + 4x + 4 = –12 + 4 (x + 5)2 = –1 (x + 2)2 = –8

28
Check It Out! Example 4a Find the zeros of the function. f(x) = x2 + 4x + 13 g(x) = x2 – 8x + 18 x2 + 4x + 13 = 0 x2 – 8x + 18 = 0 x2 + 4x = –13 + x2 – 8x = –18 + x2 + 4x + 4 = –13 + 4 x2 – 8x + 16 = – (x + 2)2 = –9 x = –2 ± 3i

29
Assignment #48 Pg 273 #2,5,6-16, 21-23, 26

30
Ch. 4.6 : I Can define and use imaginary and complex numbers and solve quadratic equations with complex roots Do Now: A farmer is fencing a yard in the shape of a trapezoid. She wants the shorter base to be 1 yard greater than the height and the longer base to be 9 yards greater than the height. She wants the area to be 200 square yards. Find the height that will give the desired area, rounded to the nearest hundredth of a yard. Success Criteria: Today’s Agenda Notes Assignment I can use complex numbers I can apply the conjugates

31
Assignment #48 Pg 273 #2,5,6-16, 21-23, 26

32
Lesson Quiz 1. Express in terms of i. Solve each equation. 3. x2 + 8x +20 = 0 2. 3x = 0 4. Find the complex conjugate of

Similar presentations

Presentation is loading. Please wait....

OK

Objectives Define and use imaginary and complex numbers.

Objectives Define and use imaginary and complex numbers.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on agriculture robotics Ppt on lcd tv technology Ppt on grease lubrication tubing Ppt on product data management Ppt on tamper resistant bolts Download ppt on software project management Ppt on varactor diode characteristics Ppt on non ferrous minerals and vitamins Ppt on the road not taken by frost Ppt on judiciary system in india