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What is the sum of all values of $k$ such that the equation $2x^2-kx+8=0$ has two distinct integer solutions?

 Mar 27, 2021
 #1
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The form of the quadratic gives us 2(x - 1)(x - 4) and 2(x - 2)(x - 2).  Since the solutions must be distinct, only 2(x - 1)(x - 4) = 2x^2 - 10x + 8 works, so k = 10.

 Mar 27, 2021
 #2
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Err, thats wrong... I used the veita.

We use the fact that the sum and product of the roots of a quadratic equation $ax^2+bx+c=0$ are given by $-b/a$ and $c/a$, respectively. Let the two roots of the equation be $p$ and $q$. Then $p+q=k/2$. However the only other restriction on $p$ and $q$ is that $pq = 4$ and that $p$ and $q$ are distinct integers. For each such possibility $(p,q)$, we also have the possibility $(-p,-q)$ since $(-p)(-q) = pq = 4$. This gives two values of $k$: $k=2(p+q)$ and $k=2(-p-q)$. Since these occur in such pairs, the sum of all possible values of $k$ is $\boxed{0}$.Alternatively, one can note that the only way to factor 4 into 2 distinct integer factors is $4\cdot1$ and $(-4)(-1)$, so that the two possible values of $k$ are $10$ and $-10$, given a sum of $0$.

 Mar 27, 2021
 #3
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Solved your own problem Lol

wolfiechan  Mar 27, 2021
 #4
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Good try though guest!

 Mar 27, 2021

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