An algebra problem by Vishal S

The roots will be in the form negative and positive, so if we add the roots, it will be 0

Let the question be in a form of $px^2 + qx + r = 0$ , we have $p=1, \space q=a, \space = 20$ .

Now, we have $r=20$ and $r$ is constant. Since the equation has the integer roots, Then, the roots will be the factors of $20$ .

For $20 = 4 \times 5$

$(x-4)(x-5) = x^2 -9x + 20 \Rightarrow a=-9$

$(x+4)(x+5) = x^2 +9x +20 \Rightarrow a=9$

For $20= 10 \times 2$

$(x-10)(x-2) = x^2 -12x +20 \Rightarrow a=-12$

$(x+10)(x+2) = x^2 + 12x +20 \Rightarrow a=12$

For $20 = 20 \times 1$

$(x-20)(x-1) = x^2 - 21x + 20 \Rightarrow a=-21$

$(x+20)(x+1) = x^2 + 21x + 20 \Rightarrow a=21$

We can clearly see that the sum of each pair of the value of $a$ is $\boxed{0}$