Multiple choice

Number of rational roots of the equation $|x^{2}-2x -3|+4x =0$ is

  1. $1$
  2. $2$
  3. $3$
  4. $4$
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A Correct answer
Explanation

Case 1: x^2 - 2x - 3 >= 0. (x-3)(x+1) >= 0, so x >= 3 or x <= -1. Equation: x^2 - 2x - 3 + 4x = 0 => x^2 + 2x - 3 = 0 => (x+3)(x-1) = 0. x = -3 (valid) or x = 1 (invalid). Case 2: x^2 - 2x - 3 < 0. -x^2 + 2x + 3 + 4x = 0 => -x^2 + 6x + 3 = 0 => x^2 - 6x - 3 = 0. Roots are (6 +/- sqrt(36 + 12))/2 = 3 +/- sqrt(12) = 3 +/- 2*sqrt(3). Neither is rational. Only x = -3 is a rational root.

AI explanation

The given equation |x^2 - 2x - 3| + 4x = 0 implies that x must be negative or zero because the absolute value is non-negative. Removing the modulus gives x^2 - 2x - 3 = -4x, which simplifies to x^2 + 2x - 3 = 0. Factoring this quadratic equation yields (x + 3)(x - 1) = 0, providing the roots x = -3 and x = 1. Since x must be negative to satisfy the initial condition, the only valid rational root is -3, making the count 1.

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