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

Question

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

Answer 1

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.

 

Answer 2

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.