If $x^{2} + 11 x + | P | = 0$ have integer roots. how many integer values can 'P' take?

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More than 8

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Solution

The correct option is D More than 8
Given equation is $x^{2} + 11 x + | p | = 0$
For the equation to have integer roots , the determinant must be greater than or equal to zero
$\Rightarrow 121 - 4 | p | \geq 0$
$| p | \leq \frac{121}{4}$
Since $p$ is integer , $| p | \leq 30$
For roots to be integers , $121 - 4 | p |$ must be a perfect square
So for $| p | = 0, 10, 18, 24, 28, 30$ , we will get integer roots
The number of values of $p$ are $11$
So the correct option is $D$

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