very typical of the discriminant case problem
\(\sqrt{11^2-4*|p|}>0\) and the discriminant actually must be
odd integer for x to have integer roots
there are not many perfect squares after the multiples of 4 are subtracted; one is \(9^2\) and \(|p|=8\)
another is \(7^2\) and \(|p|=18\)
yet, another is \(5^2\) and \(|p|=24\)
...
summing up, all perfect squares of odd numbers below 11 and including 11, i.e. 11, 9, 7, 5, 3, 1 are suitable here
p={-30, -28, -24, -18, -8, 0, 8, 18, 24, 28, 30} set contains 11 integers
answer is
A, as Quantity A=11 > Quantity B=10
KarunMendiratta wrote:
\(x^2 − 11x + |p| = 0\), where \(x\) is a variable and \(p\) is a constant, has 2 distinct integer roots.
Quantity A |
Quantity B |
Number of integer values that \(p\) can take |
10 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given