David M.

asked • 03/08/18# 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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## 2 Answers By Expert Tutors

Bobosharif S. answered • 03/08/18

Tutor

4.4
(32)
Mathematics/Statistics Tutor

In order the equation x

^{2}+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: x

_{1}=(1/2)(-11+√(121-4k)) and x_{2}=(1/2)(-11-√(121-4k)). Next, we have to select those k's for which x_{1}and x_{2}are integers. For k=10, 18, 24, 28 and 30 (which sums up exactly to 110 as you mentioned above) x_{1}and x_{2}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 x

_{1}and x_{2}.The equation x

^{2}+ 11x + k = 0 has 2 solutions ifΔ = b

^{2}- 4ac = 11

^{2}- 4k = 121 - 4k > 0

2 integer solutions requires that 121 - 4k = n

^{2}where n = 1, 2, 3, ..., 10k must be an integer requires that 4k = 121 - n

^{2}be some multiples of 4.so

n = 1, 3, 5, 7, 9

From 4k = 121 - n

^{2}k = 30, 28, 24, 18, 10

Therefore, the sum of all positive integers k such that x

^{2}+ 11x + k = 0 has 2 integer solutions is 110.## Still looking for help? Get the right answer, fast.

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Mark M.

03/08/18