6.1 B de abc-formule

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Slide 1 - Diapositive

Vandaag

Je kan de abc-formule gebruiken om alle kwadratische vergelijkingen op te lossen.

a x^{​ 2 ​ ​} + b x + c = 0

Slide 2 - Diapositive

Vandaag

Slide 3 - Diapositive

Vandaag

Waardoor kun je deze vergelijking
niet oplossen dmv ontbinden in
factoren?

Slide 4 - Diapositive

Vandaag

Slide 5 - Diapositive

De abc-formule

Van alle vergelijkingen in de vorm

zijn de oplossingen te berekenen met de abc-formule:

met

a x^{​ 2 ​ ​} + b x + c = 0

D = b^{​ 2 ​ ​} - 4 a c

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​} \lor x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

Slide 6 - Diapositive

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

x^{​ 2 ​ ​} - 4 x + 1 = 0

Slide 7 - Diapositive

a = 1, b = -4 en c = 1

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

x^{​ 2 ​ ​} - 4 x + 1 = 0

Slide 8 - Diapositive

a = 1, b = -4 en c = 1

dan de discriminant

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

x^{​ 2 ​ ​} - 4 x + 1 = 0

D = (- 4)^{​ 2 ​ ​} - 4 \cdot 1 \cdot 1

= 16 - 12 = 4

Slide 9 - Diapositive

a = 1, b = -4 en c = 1

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

x^{​ 2 ​ ​} - 4 x + 1 = 0

D = (- 4)^{​ 2 ​ ​} - 4 \cdot 1 \cdot 1 = 12

x = \frac{​ 4 + \sqrt ​ 1 2 ​ ​ ​ ​ ​}{​ 2 \cdot 1 ​} \lor x = \frac{​ 4 - \sqrt ​ 1 2 ​ ​ ​ ​ ​}{​ 2 \cdot 1 ​}

Slide 10 - Diapositive

a = 1, b = -4 en c = 1

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

x^{​ 2 ​ ​} - 4 x + 1 = 0

D = (- 4)^{​ 2 ​ ​} - 4 \cdot 1 \cdot 1 = 12

x = \frac{​ 4 + \sqrt ​ 1 2 ​ ​ ​ ​ ​}{​ 2 \cdot 1 ​} \lor x = \frac{​ 4 - \sqrt ​ 1 2 ​ ​ ​ ​ ​}{​ 2 \cdot 1 ​}

x = 3, 73 \lor x = 0, 27

Slide 11 - Diapositive

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

3 x^{​ 2 ​ ​} + 3 = 10 x

Slide 12 - Diapositive

Eerst moet de vergelijking in de goede
vorm staan (balansmethode).

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

3 x^{​ 2 ​ ​} + 3 = 10 x

3 x^{​ 2 ​ ​} - 10 x + 3 = 0

Slide 13 - Diapositive

a = 3, b = -10 en c = 3

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

3 x^{​ 2 ​ ​} - 10 x + 3 = 0

Slide 14 - Diapositive

a = 3, b = -10 en c = 3

dan de discriminant

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

D = (- 10)^{​ 2 ​ ​} - 4 \cdot 3 \cdot 3

= 100 - 36 = 64

3 x^{​ 2 ​ ​} - 10 x + 3 = 0

Slide 15 - Diapositive

a = 3, b = -10 en c = 3

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

D = (- 10)^{​ 2 ​ ​} - 4 \cdot 3 \cdot 3 = 64

3 x^{​ 2 ​ ​} - 10 x + 3 = 0

x = \frac{​ 1 0 + \sqrt ​ 6 4 ​ ​ ​ ​ ​}{​ 2 \cdot 3 ​} \lor x = \frac{​ 1 0 - \sqrt ​ 6 4 ​ ​ ​ ​ ​}{​ 2 \cdot 3 ​}

Slide 16 - Diapositive

a = 3, b = -10 en c = 3

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

D = (- 10)^{​ 2 ​ ​} - 4 \cdot 3 \cdot 3 = 64

3 x^{​ 2 ​ ​} - 10 x + 3 = 0

x = \frac{​ 1 0 + \sqrt ​ 6 4 ​ ​ ​ ​ ​}{​ 2 \cdot 3 ​} \lor x = \frac{​ 1 0 - \sqrt ​ 6 4 ​ ​ ​ ​ ​}{​ 2 \cdot 3 ​}

x = \frac{​ 1 0 + 8 ​ ​}{​ 6 ​} \lor x = \frac{​ 1 0 - 8 ​ ​}{​ 6 ​}

Slide 17 - Diapositive

a = 3, b = -10 en c = 3

a x^{​ 2 ​ ​} + b x + c = 0

x = \frac{​ - b + \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

x = \frac{​ - b - \sqrt ​ D ​ ​ ​ ​ ​}{​ 2 a ​}

D = b^{​ 2 ​ ​} - 4 a c

D = (- 10)^{​ 2 ​ ​} - 4 \cdot 3 \cdot 3 = 64

3 x^{​ 2 ​ ​} - 10 x + 3 = 0

x = \frac{​ 1 0 + \sqrt ​ 6 4 ​ ​ ​ ​ ​}{​ 2 \cdot 3 ​} \lor x = \frac{​ 1 0 - \sqrt ​ 6 4 ​ ​ ​ ​ ​}{​ 2 \cdot 3 ​}

x = 3 \lor x = \frac{​ 1 ​ ​}{​ 3 ​}

x = \frac{​ 1 0 + 8 ​ ​}{​ 6 ​} \lor x = \frac{​ 1 0 - 8 ​ ​}{​ 6 ​}

Slide 18 - Diapositive

Geef a, b en c bij de vergelijking

2 x^{​ 2 ​ ​} - x + 1 = 0

Slide 19 - Question ouverte

Wat is de discriminant van de vergelijking ?

3 x^{​ 2 ​ ​} - 7 x - 1 = 0

D = -37

D = -61

D = 61

D = 37

Slide 20 - Quiz

Los op:

2 x^{​ 2 ​ ​} - 5 x - 7 = 0

Slide 21 - Question ouverte

H6:

2, 3, 4, 5

8, 9, 11, 12, 13

15, 17, 19, 20, 21

Slide 22 - Diapositive

6.1 B de abc-formule

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