Multiple choice

If the equation $\displaystyle 2x^{3}+3x^{2}-12x+d=0$ has 3 real and distinct roots, then $d$ belongs to

  1. $\displaystyle (-\infty ,-7)\cup (20,\infty )$
  2. $(-20, 7 )$
  3. $\displaystyle (-\infty ,-20)\cup (7,\infty )$
  4. $(20, 7)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let f(x) = 2x^3 + 3x^2 - 12x + d. For 3 distinct real roots, the local maximum and minimum must have opposite signs. f'(x) = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x+2)(x-1). Critical points at x = -2 and x = 1. f(-2) = 2(-8) + 3(4) - 12(-2) + d = -16 + 12 + 24 + d = 20 + d. f(1) = 2 + 3 - 12 + d = d - 7. For opposite signs, (20+d)(d-7) < 0, so -20 < d < 7.

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