Multiple choice

The roots of the quadratic equation mx - x2 = 12, where m is a negative integer, are p and q. If |p - q| ≥ 12, what is the greatest possible integer value of m?

  1. -7

  2. -8

  3. -13

  4. -14

  5. -15
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D Correct answer
Explanation

x^2 - mx + 12 = 0. Roots p, q satisfy p+q = m and pq = 12. |p-q| = sqrt((p+q)^2 - 4pq) = sqrt(m^2 - 48). Given sqrt(m^2 - 48) >= 12, m^2 - 48 >= 144, m^2 >= 192. Since m is negative, m <= -sqrt(192) approx -13.85. The greatest integer is -14.

AI explanation

Rearrange the given equation to x squared minus mx plus 12 equals 0. The absolute difference between the roots is given by the square root of the discriminant divided by m, so the square root of m squared minus 48 is greater than or equal to 12. Squaring both sides yields m squared minus 48 is greater than or equal to 144, which simplifies to m squared being greater than or equal to 192. Since m is a negative integer, m must be less than or equal to negative 14, making negative 14 the greatest possible integer value.

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