|
© h.hofstede (h.hofstede@hogeland.nl) |
|||
 |
|||
| 1. | a. | x3
- 3x2
= 0 x2(x - 3) = 0 x2 = 0 ∨ x - 3 = 0 x = 0 ∨ x = 3 |
|
| b. | 4x5 + 2x6
= 0 2x5(2 + x) = 0 2x5 = 0 ∨ 2 + x = 0 x = 0 ∨ x = -2 |
||
| c. | x7 = 4x5
x7 - 4x5 = 0 x5(x2 - 4) = 0 x5 = 0 ∨ x2 - 4 = 0 x = 0 ∨ x2 = 4 x = 0 ∨ x = 2 ∨ x = -2 |
||
| d. | x7
- 243x2
= 0 x2(x5 - 243) = 0 x2 = 0 ∨ x5 - 243 = 0 x = 0 ∨ x5 = 243 x = 0 ∨ x = 3 |
||
| 2. | a. | x4 + 2x2
- 15 = 0 noem x2 = p p2 + 2p - 15 = 0 (p - 3)(p + 5) = 0 p = 3 ∨ p = -5 x2 = 3 ∨ x2 = -5 x = √3 ∨ x = -√3 |
|
| b. | 9x2 + x4
+ 18 = 0 noem x2 = p p2 + 9p + 18 = 0 (p + 6)(p + 3) = 0 p = -6 ∨ p = -3 x2 = -6 ∨ x2 = -3 geen oplossing |
||
| c. | x4 + 14 = 9x2
noem
x2 = p p2 - 9p + 14 = 0 (p - 7)(p - 2) = 0 p = 7 ∨ p = 2 x2 = 7 ∨ x2 = 2 x = √7 ∨ x = -√7 ∨ x = √2 ∨ x = -√2 |
||
| d. | 3x4
- 6x2
= 144 noem x2 = p 3p2 - 6p - 144 = 0 p2 - 2p - 38 = 0 (p - 8)(p + 6) = 0 p = 8 ∨ p = -6 x2 = 8 ∨ x2 = -6 x = √8 ∨ x = -√8 |
||
| 3. | a. | x8 = 4x4
+ 12 noem x4 = p p2 - 4p - 12 = 0 (p + 2)(p - 6) = 0 p = -2 ∨ p = 6 x4 = -2 ∨ x4 = 6 x = 61/4 ∨ x = -61/4 |
|
| b. | 3x4 + x7 =
x4 + 5x7 3x4 + x7 - x4 - 5x7 = 0 2x4 - 4x7 = 0 2x4 (1 - 2x3) = 0 2x4 = 0 ∨ 1 - 2x3 = 0 x4 = 0 ∨ 2x3 = 1 x = 0 ∨ x3 = 0,5 x = 0 ∨ x = 0,51/3 = 0,79 |
||
| c. | x2√x
- 2x√x = 24√x x2√x - 2x√x - 24√x = 0 √x(x2 - 2x - 24) = 0 √x = 0 ∨ x2 - 2x - 24 = 0 x = 0 ∨ (x - 6)(x + 4) = 0 x = 0 ∨ x = 6 ∨ x = -4 Maar die -4 valt af, want √-4 bestaat niet. |
||
| d. | x8
- 5x5
+ 4x2 = 0 x2 (x6 - 5x3 + 4) = 0 x2 = 0 ∨ x6 - 5x3 + 4 = 0 noem in dit tweede deel x3 = p x = 0 ∨ p2 - 5p + 4 = 0 x = 0 ∨ (p - 4)(p - 1) = 0 x = 0 ∨ p = 4 p = 1 x = 0 ∨ x3 = 4 ∨ x3 = 1 x = 0 ∨ x = 41/3 ∨ x = 1 |
||
| e. | x7
-
x5
- 6x3 = 0 x3 (x4 - x2 - 6 ) = 0 x3 = 0 ∨ x4 - x2 - 6 = 0 noem in dit tweede deel x2 = p x = 0 ∨ p2 - p - 6 = 0 x = 0 ∨ (p - 3)(p + 2) = 0 x = 0 ∨ p = 3 ∨ p = -2 x = 0 ∨ x2 = 3 ∨ x2 = -2 x = 0 ∨ x = √3 ∨ x = -√3 |
||
| 4. | a. | x = 0 geeft
y = 44 44 = -x3 + 27x + 44 -x3 + 27x = 0 x(-x2 + 27) = 0 x = 0 ∨ -x2 + 27 = 0 x = 0 ∨ x2 = 27 x = 0 ∨ x = √27 Ú x = -√27 P en R zijn de punten (√27, 44) en (-√27, 44) De afstand daartussen is 2√27 = 10,39 |
|
| b. | (x + 4)(p + 4x
- x2
) = px + 4x2 - x3 + 4p + 16x - 4x2 = -x3 + x(p + 16) + 4p dat moet gelijk zijn aan -x3 + 27x + 44 dus moet gelden p + 16 = 27 en 4p = 44 dat klopt inderdaad beiden voor p = 11 |
||
| 5. | xA
= 1 ⇒ yA =
12 + 12 • 1-2 = 1 + 12 = 13 yB = 13 geeft: 13 = x2 + 12x-2 Vermenigvuldig met x2 : 13x2 = x4 + 12 Noem nu x2 = p: 13p = p2 + 12 p2 - 13p + 12 = 0 (p - 12)(p - 1) = 0 p = 1 ∨ p = 12 x2 = 1 ∨ x2 = 12 x = 1 ∨ x = -1 ∨ x = √12 ∨ x = -√12 Omdat x > 0 is punt B: (√12, 13) = (3.46, 13) Dus xB ≈ 3,46 |
||
| 6. | Noem de beide
rechthoekszijden x en y Dan geldt 0,5 • x • y = 24 (vanwege de oppervlakte) en x2 + y2 = 100 (vanwege de schuine zijde) Uit de eerste vergelijking volgt x = 48/y en dat kun je invullen in de tweede: (48/y)2 + y2 = 100 2304/y2 + y2 = 100 vermenigvuldig alles met y2 : 2304 + y4 = 100y2 y2 = p geeft dan p2 - 100p + 2304 = 0 De ABC-formule geeft p = 64 p = 36 y2 = 64 geeft zijde y = 8 en x = 48/8 = 6 y2 = 36 geeft zijde y = 6 en x = 48/6 = 8 De zijden zijn dus 6 en 8, en dan is de omtrek 6 + 8 + 10 = 24 |
||