Sum The Integer Roots

$x^2-bx+80=0$

By Vieta's Formula , if the roots of the equation are $p$ and $q$ , then $p+q=b$ and $pq=80$ .

$\begin{aligned} 80 &=& 1\times 80 \nonumber \\ &=& 2\times 40 \nonumber \\ &=& 4\times 20 \nonumber \\ &=& 5\times 16 \nonumber \\ &=& 8\times 10 \nonumber \\ \end{aligned}$

Hence, the possible distinct values for $b$ are:

$\begin{aligned} b~~ =~~ p+q ~~&=& 1+80&=&\fbox{81} \nonumber \\ &=&2+40&=&\fbox{42} \nonumber \\ &=& 4+20&=& \fbox{24} \nonumber \\ &=& 5+16&=&\fbox{21} \nonumber \\ &=& 8+10&=&\fbox{18}\nonumber \\ \end{aligned}$

The sum is simply:

$81+42+24+21+18=\fbox{186}$

An integer greater than 0 such as $b$ can be expressed as the sum of two other integers, and never as the sum of any non-integers/integer & non-integer combination. Let us call the two integer roots of this polynomial $p$ and $q$ . Now, using common algebraic knowledge or previously acquired knowledge of Vieta's famous formulas , we know that

$p \times q = 80$

$p + q = b$ $(b>0)$

If $p \times q$ has to be positive, then we can be sure that both $p$ and $q$ have to be either both positive or both negative, but when we also consider that $p + q$ has to be positive, then we can eliminate all the possibilities in which $p$ and $q$ are both negative.

So now we just need to write out all of the positive integer factors of $80$ and them up!

$1 \times 80$

$2 \times 40$

$4 \times 20$

$5 \times 16$

$8 \times 10$

$(1 + 80) + (2 + 40) + (5+16) + (8 + 10) = \boxed{186}$