Unit 10+ Complex Numbers Lesson 30 A Little History of Complex Numbers 759

NAME: PERIOD: DATE:

Homework Problem Set

1. Determine the number and type of each solution for the following quadratic equations.

A. x2 2 6x 1 8 5 0 B. x2 2 8x 1 16 5 0 C. 4x2 1 1 5 0

2. Give a new example of a quadratic equation in standard form that has. . .

A. Exactly two distinct real solutions.

B. Exactly one distinct real solution.

C. Exactly two complex (non-real) solutions.

3. Suppose we have a quadratic equation ax2 1 bx 1 c 5 0 so that a 1 c 5 0. Does the quadratic equation have one solution or two distinct solutions? Are they real or complex? Explain how you know.

4. Write a quadratic equation in standard form such that 25 is its only solution.

bZ4aInbZ4ac 6544118 bZ4ac 854046 b24ac 024641

36 32 4 64 64 0 0 16 16TWOREALSOLUTIONS ONEREALSOLUTION TWO Complex solutions

f xS X2 16

fix x2110 25

flex x2t9

55 0710 25 0

760 Module 4 Quadratic Functions

5. Is it possible that the quadratic equation ax2 1 bx 1 c 5 0 has a positive real solution if a, b, and c are all positive real numbers?

A. What are the two solutions to the quadratic equation ax2 1 bx 1 c 5 0?

B. When will these solutions be positive?

6. Is it possible that the quadratic equation ax2 1 bx 1 c 5 0 has a positive real solution if a, b, and c are all negative real numbers? Explain your thinking.

Solve.

7. 2x2 1 8 5 0 8. x2 1 5x 1 12 5 0

9. 4x2 2 2x 1 2 5 0 10. x2 1 9 5 0

NO

btTb4aT b FacZa Za

f b ispositivethe secondwill benegativeIf btb2_4aT 0 then Iac b Therefore ifall3coefficients

SO bZ4ac b2and 4ac oathiscrenaunsstaffnonffasitiffyptheniuthereutionut

benegative

No if a band c are all negativethen a b and c are allpositiveThesolutions of ax2tbxtc 0 are thesame as solutions toar bx O l axIb tC Nopositivereal solutions

sinceall positivecoefficients

2 218 0 51 44342 5in2 2 8 2XE YI 5tzF23 5tzrt.ru

C2 I 9 FF214

2 437 iIF9 Ta Fi

2tg28IE y2jgiF I i 3i 3 i

Hart Interactive – Honors Algebra 1 M10 + HONORS ALGEBRA 1

Lesson 30 T

Homework Problem Set Sample Solutions 1. A. 𝑥𝑥2 − 6𝑥𝑥 + 8 = 0

𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 =

(−6)2 − 4(1)(8) =

36− 32 = 4

4 > 0

Two real solutions

B. 𝑥𝑥2 − 8𝑥𝑥 + 16 = 0

𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 (−8)2 − 4(1)(16) =

64− 64 = 0

One real solution

C. 4𝑥𝑥2 + 1 = 0

𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 =

02 − 4(4)(1) =

0− 16 =

−16 < 0

Two complex

solutions

2. Answers will vary.

Have students check each other’s equations. Ask how they are able to write an equation – What did they do to create their equations?

3. If 𝑎𝑎 + 𝑐𝑐 = 0, then either 𝑎𝑎 = 𝑐𝑐 = 0, a > 0 and 𝑐𝑐 < 0, or 𝑎𝑎 < 0 and 𝑐𝑐 > 0.

• The definition of a quadratic polynomial requires that 𝒂𝒂 ≠ 𝟏𝟏, so either 𝒂𝒂 > 𝟏𝟏 and 𝒄𝒄 < 𝟏𝟏

or 𝒂𝒂 < 𝟏𝟏 and 𝒄𝒄 > 𝟏𝟏.

• In either case, 𝟒𝟒𝒂𝒂𝒄𝒄 < 𝟏𝟏. Because 𝒃𝒃𝟐𝟐 is positive and 𝟒𝟒𝒂𝒂𝒄𝒄 is negative, we know 𝒃𝒃𝟐𝟐 − 𝟒𝟒𝒂𝒂𝒄𝒄 > 𝟏𝟏.

• Therefore, a quadratic equation 𝒂𝒂𝟒𝟒𝟐𝟐 + 𝒃𝒃𝟒𝟒+ 𝒄𝒄 = 𝟏𝟏 always has two distinct real

solutions when 𝒂𝒂+ 𝒄𝒄 = 𝟏𝟏.

4. (𝟒𝟒+ 𝟓𝟓)𝟐𝟐 = 𝟏𝟏 𝟒𝟒𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟒𝟒+ 𝟐𝟐𝟓𝟓 = 𝟏𝟏

Lesson 30: A Little History of Complex Numbers Module 10+: Complex Numbers

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