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JAMB 1998 · UME · Q13

Find the range of values of mm for which the roots of the equation 3x2−3mx+(m2−m−3)=03x^2 - 3mx + (m^2 - m - 3) = 0 are real.

Worked solution (try it first)
  1. Real roots need b2−4ac≥0b^2 - 4ac \ge 0: 9m2−12(m2−m−3)≥09m^2 - 12(m^2 - m - 3) \ge 0.
  2. Simplify: −3m2+12m+36≥0-3m^2 + 12m + 36 \ge 0.
  3. Divide by −3-3 and flip the sign: m2−4m−12≤0m^2 - 4m - 12 \le 0.
  4. Factorise: (m−6)(m+2)≤0(m - 6)(m + 2) \le 0, which holds between the roots.
  5. So mm lies between −2-2 and 6, option B.

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