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Difficulty: Very hardRational and Radical Expressions and Equations
For all real values of xx where the expression is defined, consider the equation:
2xx+3=x+3−1\frac{2x}{\sqrt{x + 3}} = \sqrt{x + 3} - 1
Which of the following represents the complete set of real solutions to this equation?
  1. {1}\{1\}Answer
  2. B
    {1,6}\{1, 6\}
  3. C
    {6}\{6\}
  4. D
    {−3,4}\{-3, 4\}
  5. E
    {2}\{2\}

Answer

The set containing only 1
The correct answer is the set containing only 1. To solve the equation, we first multiply both sides by the denominator x+3\sqrt{x+3}, which yields 2x=x+3−x+32x = x + 3 - \sqrt{x+3}. Isolating the radical gives x+3=3−x\sqrt{x+3} = 3-x. Squaring both sides produces the quadratic equation x+3=x2−6x+9x+3 = x^2-6x+9, which simplifies to x2−7x+6=0x^2-7x+6=0. Solving this quadratic gives candidate solutions of 1 and 6. Substituting 6 back into the original equation results in an invalid statement (4=24 = 2), making it extraneous. Substituting 1 yields a valid statement (1=11 = 1), meaning the only real solution is 1.

Step-by-Step Solution

1
Determine the domain of the equation.
x>−3x > -3
The expression inside the square root must be non-negative (x+3≥0x+3 \ge 0), and since it is in the denominator, it cannot be zero (x+3≠0x+3 \neq 0).
2
Clear the denominator by multiplying both sides by x+3\sqrt{x + 3}.
2x=x+3−x+32x = x + 3 - \sqrt{x + 3}
Multiplying both sides by the denominator simplifies the rational expression into a form where we can isolate the radical.
3
Isolate the radical term.
x+3=3−x\sqrt{x + 3} = 3 - x
Grouping all non-radical terms on one side prepares the equation for squaring to eliminate the radical.
4
Analyze constraints on the variable.
x≤3x \le 3
Since the principal square root on the left side is non-negative, the right side 3−x3 - x must also be non-negative, which restricts any valid solutions to x≤3x \le 3.
5
Square both sides and simplify to form a quadratic equation.
x2−7x+6=0x^2 - 7x + 6 = 0
Squaring both sides eliminates the radical: x+3=(3−x)2  ⟹  x+3=x2−6x+9  ⟹  x2−7x+6=0x + 3 = (3 - x)^2 \implies x + 3 = x^2 - 6x + 9 \implies x^2 - 7x + 6 = 0.
6
Solve the quadratic equation by factoring.
x=1x = 1 or x=6x = 6
Factoring (x−1)(x−6)=0(x-1)(x-6) = 0 gives the potential solutions.
7
Verify the solutions in the original equation.
x=1x = 1 is valid; x=6x = 6 is extraneous.
Substituting x=6x = 6 yields 123=4\frac{12}{3} = 4 on the left and 9−1=2\sqrt{9}-1 = 2 on the right, which are not equal. Substituting x=1x = 1 yields 22=1\frac{2}{2} = 1 on the left and 4−1=1\sqrt{4}-1 = 1 on the right, which are equal.

Key Concept

Solving equations containing both rational and radical expressions requires clearing denominators, isolating the radical, squaring both sides, and verifying candidate solutions to eliminate extraneous solutions.
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Topics

Rational and Radical Expressions and Equations

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