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Need help on this question:

If the roots of the quadratic equation $x^2 + kx - 18 = 0$ are integers, how many possible values of $k$ are there?

Know it might be something to do with the discriminant, but can't figure out exactly what to do?

root
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user256670
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  • HINT: The product of the *integral* roots is -18. That leaves only a few possibilities. – najayaz Sep 23 '15 at 17:03
  • @G-man The product of any roots must be -18. Find all possible pairs of integers that multiply to make -18. – tomi Sep 23 '15 at 17:11
  • @tomi I know that, I was just trying to phrase it in a way so as to emphasize the integer thing. – najayaz Sep 23 '15 at 17:14
  • @G-man Yes, I assumed you knew that, but wanted to make the point for the benefit of someone reading this at a later stage who might otherwise be unsure whether the product of roots property was true for all roots - integer, rational, irrational, complex, etc etc. – tomi Sep 23 '15 at 17:20

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Note that $(x-a)(x-b)=x^2-(a+b)x+ab$ so we know $ab=-18$ and furthermore $a,b$ are integers.

The possible choices for $a,b$ are (note order doesn't matter):

$a=1, b=-18$ which gives $k=17$

$a=2, b=-9$ which gives $k=7$

$a=3, b=-6$ which gives $k=3$

$a=-1, b=18$ which gives $k=-3$

$a=-2, b=9$ which gives $k=-7$

$a=-3, b=6$ which gives $k=-17$.

Matt B
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