Chapter 10 Sec 3 Completing the Square. 2 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Use Square Root...

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Chapter 10 Sec 3 Chapter 10 Sec 3 Completing the Completing the Square Square

Transcript of Chapter 10 Sec 3 Completing the Square. 2 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Use Square Root...

Page 1: Chapter 10 Sec 3 Completing the Square. 2 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Use Square Root Property Solve x 2 + 10x + 25 = 49. First & Last term.

Chapter 10 Sec 3Chapter 10 Sec 3

Completing the Completing the SquareSquare

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Algebra 1 Chapter 10 Sections 3 & 4

Use Square Root PropertyUse Square Root Property

Solve Solve xx22 + 10+ 10x x + 25 = 49.+ 25 = 49.

First & Last term perfect Squares? First & Last term perfect Squares?

Middle term = 2 Middle term = 2 xx x x x x 5?5?

Then (xThen (x + 5)(+ 5)(xx + 5) = + 5) = xx22 + 10 + 10xx + 25 + 25

So, (So, (x x + 5)+ 5)22 = 49 = 49

Sq Rt Prop. Sq Rt Prop.

x x + 5 = 7 and + 5 = 7 and x x + 5 = –7 + 5 = –7

x x = 2= 2 and x and x = –12 {2, –12} = –12 {2, –12}

7495 x

YesYes

YesYes

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Algebra 1 Chapter 10 Sections 3 & 4

Completing the SquareCompleting the Square

To solve the previous equation, the To solve the previous equation, the quadratic expression on one side quadratic expression on one side MUST MUST be a perfect square. be a perfect square.

However few quadratic expressions are However few quadratic expressions are perfect squares. perfect squares.

So we will use the following method to So we will use the following method to make a perfect square. make a perfect square.

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Algebra 1 Chapter 10 Sections 3 & 4

Completing the SquareCompleting the Square

1.

2. c = .

3. The trinomial now factors to

2

2

b

2

and , 2

, Find2

bbb

xx22 + bx + bx + + cc

2

2

bx

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Algebra 1 Chapter 10 Sections 3 & 4

Complete the SquareComplete the SquareFind the value of Find the value of c c that makes that makes xx22 + + 1212x + cx + c, then , then factor.factor.

xx22 + 12 + 12xx + 36, + 36, c = 36c = 36 ((xx + 6) + 6)22

1.1. Find Find b = , b/2 = …b = , b/2 = …

2.2. c = (b/2)c = (b/2)22

3.3. Factor to (Factor to (xx + b/2) + b/2)22

362

6,2

12

2 , 12

2

bbb

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Algebra 1 Chapter 10 Sections 3 & 4

Solving by Completing the SquareSolving by Completing the Square

1. Put all x’s on one side and all numbers on the other.

2. If a = 1, divide each term by a

3.

4. Add to both sides of the equation.

5. The factor trinomial to

6. Use the Square Root Property

7. Solve for each root, both the ( + ) and the ( – )

2

2

b

2

b and ,

2

b , Find

2

b

xx22 + bx + bx

2

2

b x

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Algebra 1 Chapter 10 Sections 3 & 4

Solve by Completing the SquareSolve by Completing the Square

Solve Solve xx22 + + 8 8x – x – 2020 = = 00 + 20 + 20+ 20 + 20 xx22 + + 8 8x = x = 2020

xx22 + + 8 8x x + 16+ 16 = 20 = 20 + 16 + 16((xx + 4) + 4)22 = 36 = 36

x = x = – 4 + 6 = 2– 4 + 6 = 2 x x = – 4 – 6 = – 10= – 4 – 6 = – 10{– 10, 2} {– 10, 2}

162

,42

8

2 ,8

2

bbb

64 x 64 x

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Algebra 1 Chapter 10 Sections 3 & 4Solve by Completing the Solve by Completing the SquareSquare

Solve Solve xx22 – – 1414x + x + 33 = = – 10 – 10 – – 3 – 3 3 – 3 xx22 – – 1414x = x = – 13– 13xx22 – – 1414x x + 49+ 49 = – 13 = – 13 + 49+ 49((xx – 7) – 7)22 = 36 = 36

b 14, b

2 14

2 7,

b

2

2

49

367 x

67 x 13

67

x

x1

67

x

x

{1, 13}

Page 9: Chapter 10 Sec 3 Completing the Square. 2 of 19 Algebra 1 Chapter 10 Sections 3 & 4 Use Square Root Property Solve x 2 + 10x + 25 = 49. First & Last term.

Chapter 10 Sec 4Chapter 10 Sec 4

Solving by Quadratic Solving by Quadratic FormulaFormula

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Algebra 1 Chapter 10 Sections 3 & 4

Quadratic FormulaQuadratic Formula

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Algebra 1 Chapter 10 Sections 3 & 4

Integral RootsIntegral Roots

xx22 – 2 – 2 x x – 24 = 0– 24 = 0

((xx + 4)( + 4)(x – x – 6) = 06) = 0

xx + 4 = 0 + 4 = 0 x x – 6 = 0– 6 = 0

xx = – 4 = – 4 x x = 6= 6

Use two methods to solve Use two methods to solve xx2 2 – 2– 2x x –– 24 = 0. 24 = 0. First, factoring.First, factoring.

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Algebra 1 Chapter 10 Sections 3 & 4

Two Rational RootsTwo Rational RootsSecond, Quadratic Equation Second, Quadratic Equation xx2 2 – 2– 2x x –– 24 = 0. 24 = 0. a a = 1= 1, b = , b = –2, –2, cc = – 24 = – 24

a

acbbx

2

42

x 2 2 2 4 1 24

2 1

2 4 962

2 1002

2 102

2 102

82

4

2 10

212

26

{– 4, 6} {– 4, 6}

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Algebra 1 Chapter 10 Sections 3 & 4

One Rational RootsOne Rational Rootsa = a = 11 , b = , b = 2222 , c = , c = 121 121 Solve Solve xx2 2 + 22+ 22x +x +121 = 0. 121 = 0.

IdentifyIdentify a, b, a, b, and and c. c.

a

acbbx

2

42

12

121142222 2 x

2

48448422 x

2

022

2

22 {– 11} {– 11}

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Algebra 1 Chapter 10 Sections 3 & 4

Two Irrational RootsTwo Irrational Roots

a = a = 2424 , b = , b = –14–14 , c = , c = – 6 – 6

Solve 24Solve 24xx2 2 – – 1414x x = 6 . = 6 . Rewrite in standard formRewrite in standard form2424xx22 – – 1414xx – – 6 = 0.6 = 0.

a

acbbx

2

42

242

62441414 2 x

x 14 196 57648

14 77248 48

77214 48

77214

14 772

48

0.3

0.9

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Algebra 1 Chapter 10 Sections 3 & 4

DiscriminantDiscriminant

a

acbbx

2

42

The The discriminantdiscriminant can be used to determine the can be used to determine the number and type of roots. The expression number and type of roots. The expression bb22 – 4 – 4ac ac is is the discriminant.the discriminant.

acb 42

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Algebra 1 Chapter 10 Sections 3 & 4

Discriminant bDiscriminant b22 – 4ac . – 4ac .

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Algebra 1 Chapter 10 Sections 3 & 4

Use the DiscriminantUse the Discriminant

a = a = 22 , b = , b = 1010 , c = , c = 11 11

State the value of the discriminant. Then determine the State the value of the discriminant. Then determine the number of real roots of the equation.number of real roots of the equation.

a = a = 44 , b = , b = – 20– 20 , c = , c = 25 25 a. 2a. 2xx22 + 10 + 10x + x + 11 = 011 = 0 b. 4b. 4xx22 – 20 – 20x + x + 25 = 025 = 0

bb22 – 4 – 4acac

(– 20)(– 20)22 – 4(4)(25) – 4(4)(25)

400 – 400 = 0 400 – 400 = 0

1 Real Root1 Real Root

(10)(10)22 – – 4(2)(11) 4(2)(11)

100 – 88 = 12100 – 88 = 12

positivepositive

2 Real Roots2 Real Roots

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Algebra 1 Chapter 10 Sections 3 & 4

Use the DiscriminantUse the Discriminant

a = 3 , b = a = 3 , b = 44 , c = , c = 2 2

State the value of the discriminant. Then determine the State the value of the discriminant. Then determine the number of real roots of the equation.number of real roots of the equation.

c. 3c. 3xx22 + 4 + 4x x = – 2= – 2Rewrite: 3Rewrite: 3xx22 + 4 + 4x x + 2 = 0+ 2 = 0

bb22 – 4 – 4acac (4)(4)22 – 4(3)(2) – 4(3)(2)

16 – 24 = – 816 – 24 = – 8

NegativeNegative

No Real RootsNo Real Roots

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Algebra 1 Chapter 10 Sections 3 & 4

Daily AssignmentDaily Assignment

• Chapter 10 Sections 3 & 4Chapter 10 Sections 3 & 4• Study Guide (SG)Study Guide (SG)

• Pg 135 - 138 OddPg 135 - 138 Odd