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√(2x - 3) + √(2x + 3) = 3 + √3 Squaring on both sides, {√(2x - 3) + √(2x + 3)}2 = (3 + √3)2 or 2x - 3 + 2x + 3 + 2 √{(2x)2 - (3) 2} = 9 + 3 + 2 x 3 √3 `` ` ` or 4x + 2 √{4x2 - 9 2} = 12 + 6 √3 So by comparing the constant part of LHS & RHS, 4x = 12 or x = 3
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