Multiple choice

The set of values of k for which the given quadratic equation has real roots $2x^2$ + kx + 3 = 0 is k $\leq -2\sqrt6$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 2x^2 + kx + 3 = 0 to have real roots, D = k^2 - 4(2)(3) >= 0. k^2 - 24 >= 0. k^2 >= 24. This means k >= sqrt(24) or k <= -sqrt(24). sqrt(24) = 2*sqrt(6). So k >= 2*sqrt(6) or k <= -2*sqrt(6). The statement says k <= -2*sqrt(6), which is only part of the solution set.

AI explanation

For real roots, the discriminant must be greater than or equal to zero, so k^2 - 4(2)(3) >= 0. This simplifies to k^2 >= 24, meaning k >= 2*sqrt(6) or k <= -2*sqrt(6). The given condition of only k <= -2*sqrt(6) is incomplete; the statement is false.

Keep practicing — related questions