Multiple choice

If the quadratic polynomial $P(x) = (p - 3)x^{2} - 2px + 3p - 6$ range from $[0, \infty)$ for every $x\epsilon R$, then the value of $p$ can be

  1. $\dfrac {3}{2}$
  2. $4$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a quadratic P(x) = ax^2 + bx + c to be >= 0 for all x, we need a > 0 and discriminant D <= 0. Here a = p - 3, so p > 3. D = (-2p)^2 - 4(p - 3)(3p - 6) = 4p^2 - 4(3p^2 - 6p - 9p + 18) = 4p^2 - 12p^2 + 60p - 72 = -8p^2 + 60p - 72. Setting D <= 0: -8p^2 + 60p - 72 <= 0, or 2p^2 - 15p + 18 >= 0. Factoring: (2p - 3)(p - 6) >= 0. This holds for p <= 1.5 or p >= 6. Combining with p > 3, we get p >= 6.

AI explanation

For the quadratic polynomial P(x) = (p - 3)x^2 - 2px + 3p - 6 to have a range of [0, infinity), the coefficient of x^2 must be positive and the discriminant must be less than or equal to zero. Applying the discriminant condition of a quadratic equation, (-2p)^2 - 4(p - 3)(3p - 6) <= 0, which simplifies to p^2 - 9p + 18 <= 0. Factoring this inequality gives (p - 3)(p - 6) <= 0, meaning p must lie in the interval [3, 6]. Since the leading coefficient (p - 3) must be strictly greater than zero, p cannot be 3, so testing the valid boundary value p = 6 satisfies all conditions.

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