Multiple choice

What is the one of the roots of the equation $\sqrt {\dfrac {2x}{3 - x}} - \sqrt {\dfrac {3 - x}{2x}} = \dfrac {3}{2}$?

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let y = sqrt(2x / (3-x)). The equation becomes y - 1/y = 3/2. Multiplying by 2y gives 2y^2 - 3y - 2 = 0, which factors to (2y + 1)(y - 2) = 0. Since y must be positive, y = 2, so 2x / (3-x) = 4, leading to 2x = 12 - 4x, or 6x = 12, so x = 2.

AI explanation

Multiplying the entire equation by sqrt(2x) * sqrt(3 - x) eliminates the fractions, giving 2x - (3 - x) = (3/2) * sqrt(2x) * sqrt(3 - x). Simplifying the left side results in 3x - 3 = (3/2) * sqrt(2x) * sqrt(3 - x), and dividing by 3 gives x - 1 = (1/2) * sqrt(6x - 2x^2). Squaring both sides produces (x - 1)^2 = (1/4)(6x - 2x^2), which expands to 4x^2 - 8x + 4 = 6x - 2x^2. Rearranging terms gives the quadratic equation 6x^2 - 14x + 4 = 0, and dividing by 2 gives 3x^2 - 7x + 2 = 0. Factoring this yields (3x - 1)(x - 2) = 0, so the valid root is 2.

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