Multiple choice

Find the values of $k$ for the following quadratic equation, so that they have two real and equal roots: $2x^2 + k x + 3 = 0$

  1. $k = \pm 2\sqrt 3$
  2. $k = \pm 2\sqrt 6$
  3. $k = \pm \sqrt 6$
  4. $k = \pm \sqrt 3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For equal roots, discriminant D = b^2 - 4ac = 0. k^2 - 4(2)(3) = 0. k^2 - 24 = 0. k^2 = 24. k = +/- sqrt(24) = +/- 2 * sqrt(6).

AI explanation

For a quadratic equation to have real and equal roots, the discriminant (b^2 - 4ac) must equal zero. Substituting a = 2, b = k, and c = 3 gives k^2 - 4(2)(3) = 0. This simplifies to k^2 - 24 = 0, so k^2 = 24. Taking the square root of both sides gives k = plus or minus 2 root 6.