1) Ö(2x - 3) = 5
Solution:
Square both sides:
Ö(2x - 3) 2 = 5 2
2x - 3 = 25
2x = 28
x = 14
Check:
Ö(2(14) - 3) = Ö(28 - 3) = Ö(25) = 5
2) Ö(x + 3) + Öx = 5
Solution:
Ö(x + 3) = 5 - Öx
Þ
x + 3 = 25 - 10 Öx + x
Simplify:
10 Öx = 22 or 5 Öx = 11
Square both sides:
25x = 121
x =121/25 = 4 21/25
Check:
Ö(121/25 + 3) + Ö(121/25) = 5
Ö(121/25 + 3) = 2.8
Ö(121/25) = 2.2
2.8 + 2.2 = 5
4) Ö(x + 5) = 1 + Öx
Solution:
Square both sides:
(Ö(x + 5))2 = (1 + Öx ) 2
The RHS will be a trinomial that has a radical as one of its terms, i.e.
x + 5 = 1 + 2 Öx + x
Simplify and isolate the radical:
4 = 2 Öx
square both sides:
(4)2 = ( 2Öx) 2
Þ
16 = 4 x
x = 4
Check:
Ö(4 + 5) = 1 + Ö4
Ö(9) = 1 + Ö4
3 = 1 + 2 = 3
5) Ö(x + 1) + Ö(2x + 3) - Ö(8x + 1) = 0
Solution:
Isolating Ö(8x + 1) :
Ö(8x + 1) = Ö(x + 1) + Ö(2x + 3)
Equating the squares of the members
x + 1 + 2 [Ö(x + 1) + Ö(2x + 3)] + 2x + 3 = 8x + 1
Isolating: 2 [Ö(x + 1) + Ö(2x + 3)]
2 [Ö(x + 1) + Ö(2x + 3)] = 8x + 1 - x - 1 - 2x - 3
Simplify radicand and combine similar terms:
2 Ö(2x 2 + 5x + 3) = 5x - 3
Squaring each side:
4 (2x 2 + 5x + 3) = 25 x 2 - 30x + 9
Apply the distributive law to LHS:
8x 2 + 20x + 12 = 25x 2 - 30x + 9
Add 25 x 2 - 30x + 9 to each side and divide by (-1):
17x 2 - 50x - 3 = 0
Apply the quadratic formula:
x = -b + Ö {b2 - 4ac}/ 2a
x = 50 + Ö(2500 + 204) / 34
x = 50 + Ö(2704) / 34
x1 = 102/ 34, x2 = - 2/ 34
x1 = 3, x2 = -1/17
Check: Substituting x1 = 3 for x in original eqn.:
Ö(3 + 1) + Ö(2(3) + 3) - Ö(8(3) + 1) = 0
Ö(4) + Ö(9) - Ö(25) = 0
2 + 3 - 5 = 0
Next subst. x2 = - 1/17 in original eqn.
Ö(-1/17 + 1) + Ö(-2/17 + 3) - Ö(- 8/17 + 1) = 0
Ö(16/17) + Ö(49/17) - Ö(9/17) = 8/Ö(17) ≠ 0
Since 8/Ö(17) ≠ 0 then x2 is not a root.
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