Multiple choice

If the equation $2x^2-6x+p=0$ has real and different roots, then the values of $p$ are given by

  1. $p<\cfrac{9}{2}$
  2. $p\leq \cfrac{9}{2}$
  3. $p>\cfrac{9}{2}$
  4. $p\geq \cfrac{9}{2}$
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A Correct answer
Explanation

For real and different roots, the discriminant D > 0. D = (-6)^2 - 4(2)(p) = 36 - 8p. 36 - 8p > 0 implies 8p < 36, so p < 36/8 = 9/2.

AI explanation

For a quadratic equation ax^2 + bx + c = 0 to have real and different roots, the discriminant must be strictly greater than zero. Applying this to 2x^2 - 6x + p = 0, we set the discriminant greater than zero, yielding (-6)^2 - 4(2)(p) > 0. Simplifying this inequality gives 36 - 8p > 0, which further reduces to 8p < 36. Dividing by 8 shows that p must be less than 9/2.

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