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√2x+3 - √x+1 = 1
Isolate one of the radicals:
√2x+3 = 1-√x+1
Square both sides
(√2x+3)² = (1-√x+1)²
Squaring the left side takes away the radical
Squaring the right side means to write the
(expression twice
2x+3 = (1-√x+1)(1-√x+1)
FOIL out the right side:
2x+3 = 1-√x+1-√x+1+(√x+1)²
Simplify
2x+3 = 1-2√x+1+x+1
2x+3 = 2+x-2√x+1
Isolate the radical term
x+1 = -2√x+1
Square both sides:
(x+1)² = (-2√x+1)²
(x+1)(x+1) = (-2)²(√x+1)²
x²+x+x+1 = 4(x+1)
x²+2x+1 = 4x+4
Get 0 on the right:
x²-2x-3 = 0
Factor the left side:
(x-3)(x+1) = 0
Us zero-factor property:
x-3 = 0; x+1 = 0
x = 3; x = -1
We must check for extraneous solutions:
Checking x = 3
√2x+3 - √x+1 = 1
√2(3)+3 - √(3)+1 = 1
√6+3 - √3+1 = 1
√9 - √4 = 1
3 - 2 = 1
1 = 1
therefore x = 3 is a solution:
Checking x = -1
√2x+3 - √x+1 = 1
√2(-1)+3 - √(1)+1 = 1
√-2+3 - √-1+1 = 1
√1 - √0 = 1
1 - 0 = 1
1 = 1
Therefore x = -1 is also a solution.
Edwin
