h.hofstede (h.hofstede@hogeland.nl)

       
1. a. √(2x - 1) - √(x - 4) = 2
(√(2x - 1) - √(x - 4))2  = 4
2x - 1 - 2 (√(2x - 1)√(x - 4)) + x - 4 = 4
2 (√(2x - 1)√(x - 4)) = 4 - 2x + 1 - x + 4
2 (√(2x - 1)√(x - 4)) = 9 - 3x
4(2x - 1)(x - 4) = (9 - 3x)2
4(2x2 - 8x - x + 4) = 81 - 54x + 9x2
8x2 - 36x + 16 = 81 - 54x + 9x2
0 = x2 - 18x + 65
0 = (x - 5)(x - 13)
x = 5 ∨  x = 13
controleren: beide oplossingen voldoen.
 
       
  b. 2√x = √(x - 3) + √(3x - 3)
4x = (√(x - 3) + √(3x - 3))2
4x = x - 3 + 2 √(x - 3)√(3x - 3) + 3x - 3
4x - x + 3 - 3x + 3 = 2 √(x - 3)√(3x - 3)
6 = 2√(x - 3)√(3x - 3)
3 = √(x - 3)√(3x - 3)
9 = (x - 3)(3x- 3)
9 = 3x2 - 9x - 3x + 9
0 = 3x2 - 12x
0 = 3x(x - 4)
x = 0   x = 4 
controleren: x = 4 voldoet.
       
  c. √(x + 7) + 2 = √(3 - x)
(√(x + 7) + 2)2 = 3 - x
x
+ 7 + 2√(x + 7)2 + 4 = 3 - x
4√(x + 7) = 3 - x - x - 7 - 4
4√(x + 7) = -2x - 8
16(x + 7) = (-2x - 8)2
16x + 112 = 4x2 + 22x8 + 64
16x + 112 = 4x2 + 32x + 64
0 = 4x2 + 16x - 48
0 = x2 + 4x  - 12
0 = (x - 2)(x + 6)
x
= 2 x = -6
controleren  x = -6 voldoet
 
       
       

h.hofstede (h.hofstede@hogeland.nl)