The equation has
- A.no solution
- B.one solution
- C.two solutions
- D.more than two solutions
Show answer and solution
Correct option: A
Solution
For the equation to be defined, we need x ≥ 1 (so both √(x+1) and √(x-1) are real) and 4x-1 ≥ 0.
Now, for x ≥ 1: √(x+1) - √(x-1) > 0, so squaring the left side gives (x+1) + (x-1) - 2√(x²-1) = 2x - 2√(x²-1).
Squaring both sides: 2x - 2√(x²-1) = 4x - 1, which gives -2√(x²-1) = 2x - 1, so 2√(x²-1) = 1 - 2x.
For x ≥ 1, the right side 1 - 2x ≤ -1 < 0, but the left side 2√(x²-1) ≥ 0. This is a contradiction for all x in the domain, so the equation has no solution.
Now, for x ≥ 1: √(x+1) - √(x-1) > 0, so squaring the left side gives (x+1) + (x-1) - 2√(x²-1) = 2x - 2√(x²-1).
Squaring both sides: 2x - 2√(x²-1) = 4x - 1, which gives -2√(x²-1) = 2x - 1, so 2√(x²-1) = 1 - 2x.
For x ≥ 1, the right side 1 - 2x ≤ -1 < 0, but the left side 2√(x²-1) ≥ 0. This is a contradiction for all x in the domain, so the equation has no solution.
[1997]