Multiple choice

The number of points $(p, q)$ such that $p, q\in \left { 1, 2, 3, 4 \right }$ and the equation $px^{2}+qx+1=0$ has real roots is

  1. $7$
  2. $8$
  3. $9$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For real roots, the discriminant D = q^2 - 4p >= 0, so q^2 >= 4p. Testing values for p, q in {1, 2, 3, 4}: If p=1, q^2 >= 4 (q=2,3,4 -> 3 pairs). If p=2, q^2 >= 8 (q=3,4 -> 2 pairs). If p=3, q^2 >= 12 (q=4 -> 1 pair). If p=4, q^2 >= 16 (q=4 -> 1 pair). Total = 3 + 2 + 1 + 1 = 7.

AI explanation

For the equation px^2 + qx + 1 = 0 to have real roots, its discriminant must be non-negative, meaning q^2 - 4p >= 0. Since p and q are chosen from the set {1, 2, 3, 4}, we count the valid ordered pairs (p, q). If p = 1, q^2 >= 4, giving q = 2, 3, or 4 for 3 pairs. If p = 2, q^2 >= 8, giving q = 3 or 4 for 2 pairs. If p = 3, q^2 >= 12, giving q = 4 for 1 pair, and if p = 4, q^2 >= 16, giving q = 4 for 1 pair. Adding these valid pairs gives a total of 3 + 2 + 1 + 1 = 7 points.

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