The product of the roots $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ is $1$. That part was easy! The sum will be more work.
The sum $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}$ of the roots simplifies to $\dfrac{\alpha^2+\beta^2}{\alpha\beta}$.
But $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$. Thus the sum of the roots is $\dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$.
Substituting the known values of $\alpha+\beta$ and $\alpha\beta$ we find that the sum of the roots is $\dfrac{(-8)^2-2(-5)}{-5}$.
This simplifies to $-\dfrac{74}{5}$.
Thus the equation is
$$x^2+\frac{74}{5}x+1=0.$$
We can multiply through by $5$ if we wish.
What we have is a line with positive slope, passing through the point $(-2, 5)$, with the length of the segment on this line from the point where it intersects the negative $x$-axis at $(x_0, 0)$ and the positive $y$-axis at $(0, y_0)$ equal to $7\sqrt 2$.
We know then, that $$x_0^2 + y_0^2 = (7\sqrt 2)^2 = 98\tag{1}$$
We also know that the distance from $(x_0, 0)$ to $(-2, 5)$ plus the distance from $(-2, 5)$ to the point $(0, y_0)$ is equal to $7\sqrt 2$: $$\sqrt {25 + (x_0+2)^2} + \sqrt{4+(y_0-5)^2} = 7\sqrt 2\tag{2}.$$
Solving $(1)$ and $(2)$ simultaneously gives us the real solutions $x_0 = -7$, and $y_0 = 7$. With these values, we must have that the slope of each of these "segments" comprising the entire length are equal: $$m = \frac{5-0}{-2 - x_0} = \frac{5-y_0}{-2-0}= 1$$
Now, given your point on the line $(-2, 5)$ and the slope of $m = 1$, we can construct the equation of the line: $$y - 5 = x + 2 \iff y = x+7$$
Best Answer
Let's say the quadratic equation is $ax^2+bx+c=0$. Without loss of generality, we assume that $a=1$. Then, the roots are
$$x_{\pm}=\frac{-b\pm\sqrt{b^2-4c}}{2}$$
The product of the roots are simply
$$x_+x_-=c$$
and since we must have $\frac12 x_+x_-=A$, then $c=2A$. Our quadratic equation is now of the form $x^2+bx+2A=0$.
We also know that the line that passes through $(0,x_+)$ and $(x_-,0)$ also passes through $(1,1)$. Thus, we have
$$\frac{x_+-1}{1}=\frac{x_+}{x_-}\implies x_+x_-=x_++x_-\implies c=-b\implies b=-c=-2A$$
Thus, the quadratic equation is
$$\bbox[5px,border:2px solid #C0A000]{x^2-2Ax+2A=0}$$