Wiskunde.net

53.
a.
x² - 25 = 0
x² = 25
x = 5 v x = -5
b.
(x - 1)(x + 3) = 0
x - 1 = 0 v x + 3 = 0
x = 1 v x = -3

54.
a.
x² + 6x = 0 (haal x voor de haakjes)
x(x + 6) = 0
x = 0 v x + 6 = 0
x = 0 v x = -6
b.
x² + 6x = 7
x² + 6x - 7 = 0 (product-som-methode met -7)
(x - 1)(x + 7) = 0
x - 1 = 0 v x + 7 = 0
x = 1 v x = -7
c.
x² + 6x + 7 = 0 (lukt niet met product-som-methode, dus abc-formule)
a = 1, b = 6 en c = 7
D = b² - 4ac = (6)² - 4 ⋅ 1 ⋅ 7 = 36 - 28 = 8
x₁ = (-6 - √8) / 2 v x₂ = (-6 + √8) / 2
x₁ ≈ -4,41 v x₂ ≈ -1,59
d.
6x² + 36x = 96
6x² + 36x - 96 = 0 (links en rechts :6)
x² + 6x - 16 = 0 (product-som-methode met -16)
(x - 2)(x + 8) = 0
x - 2 = 0 v x + 8 = 0
x = 2 v x = -8
e.
x² + 6 = 0
x² = -6
Kan niet, dus geen oplossingen. Een kwadraat kan nooit negatief zijn.
f.
-x² + 7x + 6 = 0
x² - 7x - 6 = 0 (lukt niet met product-som-methode, dus abc-formule)
a = 1, b = -7 en c = -6
D = b² - 4ac = (-7)² - 4 ⋅ 1 ⋅ -6 = 49 + 24 = 73
x₁ = (7 - √73) / 2 v x₂ = (7 + √73) / 2
x₁ ≈ -0,77 v x₂ ≈ 7,77

55.
a.
-x² + 3x + 1 = 0
x² - 3x - 1 = 0 (lukt niet met product-som-methode, dus abc-formule)
a = 1, b = -3 en c = -1
D = b² - 4ac = (-3)² - 4 ⋅ 1 ⋅ -1 = 9 + 4 = 13
x₁ = (3 - √13) / 2 v x₂ = (3 + √13) / 2
x₁ ≈ -0,30 v x₂ ≈ 3,30
b.
4x² - 8x = 0 (haal 4x voor de haakjes)
4x(x - 2) = 0
4x = 0 v x - 2 = 0
x = 0 v x = 2
c.
4x² + 4 = 8x
4x² - 8x + 4 = 0 (links en rechts :4)
x² - 2x + 1 = 0 (product-som-methode met +1)
(x - 1)(x - 1) = 0
x - 1 = 0 v x - 1 = 0
x = 1 v x = 1
d.
3x - 1 = 2x²
-2x² + 3x - 1 = 0
a = -2, b = 3 en c = -1
D = b² - 4ac = (3)² - 4 ⋅ -2 ⋅ -1 = 9 - 8 = 1
x₁ = (-3 - √1) / -4 v x₂ = (-3 + √1) / -4
x₁ = (-3 - 1) / -4 v x₂ = (-3 + 1) / -4
x₁ = -4/-4 = 1 v x₂ = -2/-4 = 1/2
e.
1/2x² + 5x = 12
1/2x² + 5x - 12 = 0 (links en rechts keer 2)
x² + 10x - 24 = 0 (product-som-methode met -24)
(x - 2)(x + 12) = 0
x - 2 = 0 v x + 12 = 0
x = 2 v x = -12
f.
(x - 2)(x - 3) = 20
x² - 3x - 2x + 6 = 20
x² - 5x - 14 = 0 (product-som-methode met -14)
(x + 2)(x - 7) = 0
x + 2 = 0 v x - 7 = 0
x = -2 v x = 7

56.
a.
(2x + 3)² = 36 (van de vorm x² = c)
2x + 3 = 6 v 2x + 3 = -6
2x = 3 v 2x = -9
x = 1½ v x = -4½
b.
(7x + 8)(3x - 57) = 0
7x + 8 = 0 v 3x - 57 = 0
7x = -8 v 3x = 57
x = -8/7 = -1 1/7 v x = 57/3 = 19
c.
1/15x² + 1/5x = 1/3 (links en rechts keer 15)
x² + 3x - 5 = 0
a = 1, b = 3 en c = -5
D = b² - 4ac = (3)² - 4 ⋅ 1 ⋅ -5 = 9 + 20 = 29
x₁ = (-3 - √29) / 2 v x₂ = (-3 + √29) / 2
x₁ ≈ -4,19 v x₂ ≈ 1,19
d.
(x - 1)(3x + 7) = -3
3x² + 7x - 3x - 7 = -3
3x² + 4x - 4 = 0
a = 3, b = 4 en c = -4
D = b² - 4ac = (4)² - 4 ⋅ 3 ⋅ -4 = 16 + 48 = 64
x₁ = (-4 - √64) / 6 v x₂ = (-4 + √64) / 6
x₁ = (-4 - 8) / 6 v x₂ = (-4 + 8) / 6
x₁ = -12/6 = -2 v x₂ = 4/6 = 2/3
e.
x(x - 1) = 1/5 - 4/5x
x² - x = 1/5 - 4/5x (links en rechts keer 5)
5x² - 5x = 1 - 4x
5x² - x - 1 = 0
a = 5, b = -1 en c = -1
D = b² - 4ac = (-1)² - 4 ⋅ 5 ⋅ -1 = 1 + 20 = 21
x₁ = (1 - √21) / 10 v x₂ = (1 + √21) / 10
x₁ ≈ -0,36 v x₂ ≈ 0,56
f.
(3x - 1)² - (2x + 1)² = 7
9x² - 6x + 1 - 4x² - 4x - 1 = 7
5x² - 10x - 7 = 0
a = 5, b = -10 en c = -7
D = b² - 4ac = (-10)² - 4 ⋅ 5 ⋅ -7 = 100 + 140 = 240
x₁ = (10 - √240) / 10 v x₂ = (10 + √240) / 10
x₁ ≈ -0,55 v x₂ ≈ 2,55

57.
a.
Los op: 0,00096x² - 0,48x + 75 = 75
0,00096x² - 0,48x = 0 (links en rechts :0,00096)
x² - 500x = 0
x(x - 500) = 0
x = 0 v x - 500 = 0
x = 0 v x = 500
Dus de pylonen staan 500 meter uit elkaar.
b.
Het laagste punt bevindt zich bij de X_top. De X_top zit in het midden tussen 0 en 500.
Dus X_top = 250.
X_top invullen in de formule levert:
y = 0,00096 ⋅ (250)² - 0,48 ⋅ 250 + 75
y = 60 - 120 + 75
y = 15
Dus het laagste punt (Y_top) is op 15 meter boven de fundering (x-as).
Het laagste punt tot de weg is 4 meter. Dus het wegdek hangt op hoogte 15 - 4 = 11 meter.

58.
oppervlakte(totaal) = lengte x breedte = (40 + 2x)(30 + 2x) = 3000
1200 + 80x + 60x + 4x² = 3000
4x² + 140x - 1800 = 0 (links en rechts :4)
x² + 35x - 450 = 0 (product-som-methode met -450)
(x - 10)(x + 45) = 0
x - 10 = 0 v x + 45 = 0
x = 10 v x = -45
x = -45 kan niet want een negatieve afmeting bestaat niet.
Dus de lijst is 10 cm breed.

59.
opp(I) = x ⋅ (3 + 2x) = 3x + 2x²
opp(II) = 15 ⋅ x = 15x
opp(III) = x ⋅ (7 + 2x) = 7x + 2x²
opp(IV) = x ⋅ (5 + x) = 5x + x²
opp(V) = 4 ⋅ x = 4x
opp(VI) = x ⋅ (10 - x) = 10x - x²
Tel nu alles op en je krijgt als totale oppervlakte: 4x² + 44x
Dus oppervlakte tegelpad = 4x² + 44x
4x² + 44x = 104 (links en rechts :4)
x² + 11x - 26 = 0 (product-som-methode met -26)
(x - 2)(x + 13) = 0
x - 2 = 0 v x + 13 = 0
x = 2 v x = -13
Dus breedte van het pad is 2 meter.

60.
We hebben dus 2 vergelijkingen:
(1) 9x + 8y = 576
(2) x ⋅ y = 1152 => y = 1152/x

Vul y = 1152/x in (1) levert:
9x + 8 ⋅ (1152/x) = 576
9x + 9216/x = 576 (links en rechts keer x)
9x² + 9216 = 576x
9x² - 576x + 9216 = 0 (links en rechts :9)
x² - 64x + 1024 = 0
a = 1, b = -64 en c = 1024
D = b² - 4ac = (-64)² - 4 ⋅ 1 ⋅ 1024 = 4096 - 4096 = 0
x₁ = (64 - √0) / 2 v x₂ = (64 + √0) / 2
x₁ = 64/2 v x₂ = 64/2
x₁ = 32 v x₂ = 32
Dus x = 32
x ⋅ y = 1152 => y = 1152/32 = 36
Dus x = 32 meter en y = 36 meter.