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What is the sum of all positive integers k such that the equation x^2 + 11x + k = 0 has two integer solutions?

The answer is 110 but I don't know the solution.

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The sum of ALL of the positive integers that solve is unlimited.
The first few are -12, -26, -42, 
Check if the conditions given are correct.

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Bobosharif S. | Mathematics/Statistics TutorMathematics/Statistics Tutor
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In order the equation x2+11x+k=0 to have two (different) solutions, it should be 121-4k>0. From here it follows that k<121/4. Since it is required k to be a positive integer, 0<k≤30=[121/4], (here [a] denotes the integer part of a)
 
For k<121/4, the two roots of the equation are: x1=(1/2)(-11+√(121-4k)) and x2=(1/2)(-11-√(121-4k)). Next, we have to select those k's for which x1 and x2 are integers.  For k=10, 18, 24, 28 and 30 (which sums up exactly to 110 as you mentioned above) x1 and x2 are Integers. The corresponding pair of solutions (roots) are: {-1, -10}, {-2, -9}, {-3, -8}, {-4, -7} , {-5, -6}
(For k=0 the solution is integers as well but 0 is not positive)
Now, the only question to be clarified is: how to find those k's which make resulting roots Integers. I assume it might be interesting for you to think about it. Hint: Analyse the expression for x1 and x2.
 
Kris V. | Experienced Mathematics, Physics, and Chemistry TutorExperienced Mathematics, Physics, and Ch...
5.0 5.0 (3 lesson ratings) (3)
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The equation x2 + 11x + k = 0 has 2 solutions if
Δ = b2 - 4ac
   = 112 - 4k
   = 121 - 4k > 0
 
2 integer solutions requires that 121 - 4k = n2 where n = 1, 2, 3, ..., 10
k must be an integer requires that 4k = 121 - n2 be some multiples of 4.
so
n = 1, 3, 5, 7, 9
 
From 4k = 121 - n2
k = 30, 28, 24, 18, 10
 
Therefore,  the sum of all positive integers k such that x2 + 11x + k = 0 has 2 integer solutions is 110.