Multiple choice

Suppose k is any integer such that the equation 2x2 + kx + 5 = 0 has no real roots and the equation x2 + (k − 5)x + 1 = 0 has two distinct real roots for x. Then, the number of possible values of k is

  1. 7

  2. 8

  3. 9

  4. 13
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

No real roots for 2x^2+kx+5=0 means discriminant D = k^2 - 40 < 0, so k^2 < 40. k is in {-6, -5, ..., 5, 6}. Two distinct real roots for x^2+(k-5)x+1=0 means D = (k-5)^2 - 4 > 0, so (k-5)^2 > 4. This means k-5 > 2 or k-5 < -2, so k > 7 or k < 3. Combining k^2 < 40 and (k < 3 or k > 7): k can be -6, -5, -4, -3, -2, -1, 0, 1, 2. That is 9 values.

AI explanation

For the equation 2x^2 + kx + 5 = 0 to have no real roots, its discriminant must be negative, meaning k^2 - 40 < 0 and -6.32 < k < 6.32. For x^2 + (k - 5)x + 1 = 0 to have two distinct real roots, the discriminant must be positive, giving (k - 5)^2 - 4 > 0, so k > 7 or k < 3. Combining these two conditions, the integer values of k must fall in the range -6.32 < k < 3, giving the nine possible values -6, -5, -4, -3, -2, -1, 0, 1, and 2.

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