Two identical conducting spheres carrying different charges attract each other . . .

Question : Two identical conducting spheres carrying different charges attract each other with a force F when placed in air medium at a distance d apart. The spheres are brought into contact and then taken to their original positions. Now, the two spheres repel each other with a force whose magnitude is equal to that of the initial attractive force. The ratio between initial charges on the spheres is 

(A) -(3+√8) only

(B) -3+√8 only 

(C) -(3+√8) OR (-3+√8)

(D) -√8

Doubt by Gurdev

Solution : 
Let the initial charges be q₁ and -q₂.

Using Coulomb's Law
F=k|q₁||-q₂|/d²
F=kq₁q₂/d² — (1) 

When the two spheres are bought into contact with each other then new charge on them would be 
q₁'=q₂'=(q₁-q₂)/2

F=k|q₁'||q₂'|/d²
F=k[(q₁-q₂)/₂][(q1-q₂)/₂]/d² — (2) 

Equating equation (1) and (2) 

kq₁q₂/d²=k[(q₁-q₂)/2][(q₁-q₂)/2]/d²
q₁q₂=(q₁-q₂)²/4
4q₁q₂=(q₁-q₂)²
(q₁-q₂)²=4q₁q₂
q₁²+q₂²-2q₁q₂=4q₁q₂
q₁²+q₂²-2q₁q₂-4q₁q₂=0
q₁²+q₂²-6q₁q₂=0

dividing both sides by q₂²
q₁²/q₂²+1-6q₁/q₂=0
(q₁/q₂)²+1-6(q₁/q₂)=0

Let q₁/q₂=x
x²+1-6x=0
x²-6x+1=0
a=1

b=-6

c=1

D=b²-4ac
D=(-6)²-4(1)(1)
D=36-4
D=32

x=[-b±√D]/2a
x=[-(-6)±√32]/2(1)
x=[6±4√2]/2
x=2[3±2√2]/2
x=3±2√2
|q₁/q₂|=3±2√2
|q₁/q₂|=3±√8
q₁/q₂=(3±√8)/(-1)
q₁/q₂=-(3±√8)
q₁/q₂=-(3+√8) or (-3+√8)
Hence, (C) -(3+√8) OR (-3+√8), would be the correct option.