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Samenvatting Getal & Ruimte 12e ed 3 vwo uitwerkingen deel 1 H3
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Getal & Ruimte 12e ed 3 vwo uitwerkingen deel 1 Hoofdstuk 3: Kwadratische vergelijkingen
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UITWERKINGENDEEL 1
WISKUNDE 12e EDITIE
3 VWO
,Getal & Ruimte
Uitwerkingen
3 vwo deel 1
Twaalfde editie, 2019 Auteurs
Noordhoff J. H. Dijkhuis
Groningen C. J. Admiraal
J. A. Verbeek
G. de Jong
H. J. Houwing
J. D. Kuis
F. ten Klooster
S. K. A. de Waal
J. van Braak
J. H. M. Liesting-Maas
M. Wieringa
M. L. M. van Maarseveen
R. D. Hiele
J. E. Romkes
M. Haneveld
S. Voets
I. Cornelisse
M. Vos
B.W. van Laarhoven
, 3 Kwadratische problemen
Voorkennis Ontbinden in factoren en vergelijkingen oplossen
Bladzijde 94
1 a x2 − 12x = x(x − 12) d 8x2 + 12x = 4x(2x + 3)
b x2 + x = x(x + 1) e 16x2 − 48x = 16x(x − 3)
c 5x2 − 3x = x(5x − 3) f 5x2 + 160x = 5x(x + 32)
2 a x2 + 8x + 12 = (x + 2)(x + 6) d x2 − x − 12 = (x + 3)(x − 4)
b x2 − 7x + 12 = (x − 3)(x − 4) e x2 + x − 12 = (x − 3)(x + 4)
c x2 − 4x − 12 = (x + 2)(x − 6) f x2 − 13x + 12 = (x − 1)(x − 12)
Bladzijde 95
3 a x2 − 7x − 18 = (x + 2)(x − 9) d x2 + x − 6 = (x − 2)(x + 3)
b x2 + 9x + 18 = (x + 3)(x + 6) e x2 − x − 2 = (x + 1)(x − 2)
c x2 + 7x + 6 = (x + 1)(x + 6) f x2 + 3x + 2 = (x + 1)(x + 2)
4 a x2 + 10x + 16 = 0 d x2 − 2x − 15 = 0 g x2 + 7x − 30 = 0
(x + 2)(x + 8) = 0 (x + 3)(x − 5) = 0 (x − 3)(x + 10) = 0
x+2=0∨x+8=0 x+3=0∨x−5=0 x − 3 = 0 ∨ x + 10 = 0
x = −2 ∨ x = −8 x = −3 ∨ x = 5 x = 3 ∨ x = −10
b x2 + 10x = 0 e (2x − 1)(3x + 1) = 0 h x2 + x − 72 = 0
x(x + 10) = 0 2x − 1 = 0 ∨ 3x + 1 = 0 (x − 8)(x + 9) = 0
x = 0 ∨ x + 10 = 0 2x = 1 ∨ 3x = −1 x−8=0∨x+9=0
x = 0 ∨ x = −10 1 1 x = 8 ∨ x = −9
x = 2 ∨ x = −3
c (x − 10)(2x + 5) = 0 i x2 + x = 0
f 2x2 − x = 0
x − 10 = 0 ∨ 2x + 5 = 0 x(x + 1) = 0
x(2x − 1) = 0
x = 10 ∨ 2x = −5 x=0∨x+1=0
1 x = 0 ∨ 2x − 1 = 0
x = 10 ∨ x = − 22 x = 0 ∨ x = −1
x = 0 ∨ 2x = 1
1
x=0∨x=2
5 a x2 + 2x = 24 c x2 − 7 = 6x e x2 = 16x
x2 + 2x − 24 = 0 x2 − 6x − 7 = 0 x2 − 16x = 0
(x − 4)(x + 6) = 0 (x + 1)(x − 7) = 0 x(x − 16) = 0
x−4=0∨x+6=0 x+1=0∨x−7=0 x = 0 ∨ x − 16 = 0
x = 4 ∨ x = −6 x = −1 ∨ x = 7 x = 0 ∨ x = 16
b x2 = 3x + 18 d x2 + 40 = 14x f x2 = 16x + 36
x2 − 3x − 18 = 0 x2 − 14x + 40 = 0 x2 − 16x − 36 = 0
(x + 3)(x − 6) = 0 (x − 4)(x − 10) = 0 (x + 2)(x − 18) = 0
x+3=0∨x−6=0 x = 4 ∨ x = 10 x + 2 = 0 ∨ x − 18 = 0
x = −3 ∨ x = 6 x = −2 ∨ x = 18
6 a x(x + 1) = 5x c (2t − 1)(t + 2) = 0 e (p − 4)(p + 4) = 6p
x2 + x = 5x 2t − 1 = 0 ∨ t + 2 = 0 p2 − 16 = 6p
x2 − 4x = 0 2t = 1 ∨ t = − 2 p2 − 6p − 16 = 0
x(x − 4) = 0 1 (p + 2)(p − 8) = 0
t = 2 ∨ t = −2
x=0∨x−4=0 p+2=0∨p−8=0
d (x + 3)2 = 2x + 9
x=0∨x=4 p = −2 ∨ p = 8
x2 + 6x + 9 = 2x + 9
b (x − 2)2 = 14 − x f (2a + 5)(2a − 5) = 0
x2 + 4x = 0
x2 − 4x + 4 = 14 − x 2a + 5 = 0 ∨ 2a − 5 = 0
x(x + 4) = 0
x2 − 3x − 10 = 0 2a = − 5 ∨ 2a = 5
x=0∨x+4=0 1 1
(x + 2)(x − 5) = 0 a = − 22 ∨ a = 22
x = 0 ∨ x = −4
x+2=0∨x−5=0
x = −2 ∨ x = 5
© Noordhoff Uitgevers bv Kwadratische problemen 69
, 3.1 Kwadratische vergelijkingen
Bladzijde 96
1 a x2 + 2x − 24 = 0
(x − 4)(x + 6) = 0
x−4=0∨x+6=0
x = 4 ∨ x = −6
b Dit heeft geen zin omdat de factor voor x2 niet gelijk is aan 1.
De product-som-methode kan alleen bij de vorm x2 + …x + … = 0.
c 2x2 + 2x − 24 = 0
x2 + x − 12 = 0
(x + 4)(x − 3) = 0
x+4=0∨x−3=0
x = −4 ∨ x = 3
1
2 a 2x2 − 32x − 72 = 0 c − x2 + 6x − 5 = 0 e − 2 x2 = 3x − 8
x2 − 16x − 36 = 0 x2 − 6x + 5 = 0 1
(x + 2)(x − 18) = 0 (x − 1)(x − 5) = 0 − 2 x2 − 3x + 8 = 0
x = − 2 ∨ x = 18 x=1∨x=5 x2 + 6x − 16 = 0
1 2 1 d 3x2 + 135 = 54x (x + 8)(x − 2) = 0
b 4 x + 32 x + 12 = 0
3x2 − 54x + 135 = 0 x = −8 ∨ x = 2
x2 + 14x + 48 = 0
x2 − 18x + 45 = 0 f 4x2 = 60 − 8x
(x + 8)(x + 6) = 0
(x − 15)(x − 3) = 0 4x2 + 8x − 60 = 0
x = −8 ∨ x = −6
x = 15 ∨ x = 3 x2 + 2x − 15 = 0
(x + 5)(x − 3) = 0
x = −5 ∨ x = 3
Bladzijde 97
1 1 1
3 a − 2x2 + 48 = 10x c (2 x − 3)2 = 12 − 4 x e (x − 2)(2x − 5) = x + 2
− 2x2 − 10x + 48 = 0 1 2 1 1 2x2 − 5x − 4x + 10 = x + 2
x2 + 5x − 24 = 0 4 x − 3x + 9 = 12 − 4 x 2x2 − 10x + 8 = 0
1 2 3 1
(x + 8)(x − 3) = 0 4 x − 24 x + 72 = 0 x2 − 5x + 4 = 0
x = −8 ∨ x = 3 (x − 1)(x − 4) = 0
x2 − 11x + 30 = 0
b (3x − 1)(5x + 30) = 0 x=1∨x=4
(x − 5)(x − 6) = 0
3x − 1 = 0 ∨ 5x + 30 = 0 f (2x − 1)2 = 12x − 11
x=5∨x=6
3x = 1 ∨ 5x = − 30 4x2 − 4x + 1 = 12x − 11
1 d (x − 5)2 = 2x − 10
x = 3 ∨ x = −6 4x2 − 16x + 12 = 0
x2 − 10x + 25 = 2x − 10
x2 − 4x + 3 = 0
x2 − 12x + 35 = 0
(x − 1)(x − 3) = 0
(x − 5)(x − 7) = 0
x=1∨x=3
x=5∨x=7
4 a Zie de figuur hiernaast.
opp I = 8 · x = 8x
2x II III II 2x
opp II = 2x · x = 2x2
opp III = 12 · 2x = 24x
12
opp tegelpad = 2 · opp I + 2 · opp II + opp III
= 2 · 8x + 2 · 2x2 + 24x
= 16x + 4x2 + 24x
I 8 8 I
= 4x2 + 40x
b 4x2 + 40x = 44
c 4x2 + 40x = 44
4x2 + 40x − 44 = 0 x x
x2 + 10x − 11 = 0
(x + 11)(x − 1) = 0
x = − 11 ∨ x = 1
De oplossing x = −11 kan niet, dus aan de zijkanten is het tegelpad 1 meter breed.
Aan de achterkant is het tegelpad 2 meter breed.
70 Hoofdstuk 3 © Noordhoff Uitgevers bv
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