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## Integer Roots

x2 - 11x + |p| = 0 has integer roots. How many integer values can 'p' take?
1. 6
2. 4
3. 8
4. More than 8

Choice D. More than 8

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## Detailed Solution

To start with the discriminant should be a perfect square. Let the dicriminant be 'D'.

From the quadratic formula: ${\frac{-11 \pm \sqrt{ D } }{2}}$, we see that the numerator has to be an even number for the roots to be integers.

This implies that the discriminant should be a perfect square and be square of an odd number. (Only then we will have odd + odd = even in the numerator)

D = 121 – 4|p| = 121 – 4|p|
4|p| cannot be negative => D can take values 121, 81, 49, 25, 9, 1
|p| can be 0, 10, 18, 24, 28, 30

p can take 0, ±10, ± 18, ± 24, ± 28, ± 30

Correct Answer: More than 8 values are possible.

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