Solve the quadratic equation by factorization method

1/x +1/(x-6)=1/4
((x-6)+x)/x^2-6x=1/4
x^2-14x +24=0
x^2 -2x -12x +24=0
x(x-2) -12(x-2)=0
(x-12)(x-2)=0
therefore either B takes 12 days and A takes 6 or B takes 2 and A takes -4
So obviously B takes 12 days

The sum of their radii must be 14 cm since that is the distance between their centres..
let their radii be r and 14-r
sum of areas is-
pi(r)^2 + pi(14-r)^2 =130pi
r^2 + 196 + r^2 -28r=130
r^2 -14r +33=0
r^2 -11r -3r +33=0
r(r-11) -3(r-11)=0
(r-3)(r-11)=0
therefore radii are 3 and 11 cm

Ques1. A TAKES 6 DAYS LESS THAN B TO FINISH A PIECE OF WORK IF BOTH A & B TOGETHER CAN FINISH IT IN 4 DAYS ,FIND THE TIME TAKEN BY B TO FINISH THE WORK.FIND ITS SOLUTION ONLY
B(B -- 6) = 4(B + B -- 6) giving B^2 -- 10B + 24 = 0 giving (B -- 12)(B + 2) = 0 OR B = 12 days ANSWER discarding --2.
Ques2. TWO CIRCLES TOUCH EXTERNALLY.THE SUM OF THEIR AREA IS 130π SQUARE CM & DISTANCE BETWEEN THEIR CENTRES IS 14 cm .
A^2 + B^2 = 130 and (A + B) = 14 whence 2AB = (A + B)^2 -- (A^2 + B^2) = 14^2 -- 130 = 66 giving
AB = 33 then A + 33/A = 14 gives A^2 -- 14A + 33 = 0 OR A = 7 +/-- 4 = 11 or 3
Radii are 11 cm and 3 cm ANSWER

For question 1, this can be VERY confusing unless you define your algebra properly.
For example, a possible definition could be:
Let a = The RATE of work accomplished by person a.
Let b = The RATE of work accomplished by person b.
Let x = The AMOUNT of work needed to do.
And so on...