Quadra-trick

Let the equation be $ax^2+bx+c=0$ .

Dividing by $a$ gives $x^2+\frac{b}{a}x+\frac{c}{a}=0$

So, according to the question $\frac{-b}{a}+\frac{c}{a}=8$

By assuming $\frac{b}{a}=p$ and $\frac{c}{a}=q$ ,

we get $x^2+px+q=0$ and $q-p=8$

Let the roots of the equation be $m$ and $n$ .

So, $mn+(m+n)=8$

$\Rightarrow mn+(m+n)+1=9$

$\Rightarrow (1+m)(1+n)=9$

9 can be factorised as

$9=3\times3=-3\times-3=9\times1=-9\times-1$ .

Since there are four unordered factorised form of 9 possible, therefore there are four equations with different roots satisfying such condition.

Let the roots are a, and b, ab+(a+b)=8, or (a+1)(b+1)=9, So (a,b)=(2,2),(-4,-4),(0,8),(8,0),(-2,-10) and (-10,-2).And there are 4 Quadratics, respectively : x^2-4x+4=0,x^2+8x+16, x^2-8x=0, and x^2+12x+20=0 There are 4 Quadratic equations