7 and 1 for the first question??

answers to Q1 are 1 and 7

Conventions used:
John = J
Kate = K
xJ1 = 6
xJ2 = 2
xK1 = -7
xK2 = -1
c' = incorrect value for C
b' = incorrect value for B

The wording tells us that:
axJ^2 + bxJ + c' = 0 (equation 1)
axK^2 + b'xK + c = 0 (equation 2)
while we need to find x1 and x2 satisfying the quadratic equation ax^2 + bx + c = 0

the first move is to express c' in terms of a & b, and b' in terms of a & c, then express b & c in terms of a

Injecting final values in equation 1 gives us:
36a + 6b + c' = 0 and 4a + 2b + c' = 0
i.e. 36a + 6b = 4a + 2b
i.e. b = -8a

Injecting final values in equation 2 gives us:
49a - 7b' + c = 0 and a - b' + c = 0
i.e. 7a + c/7 = a + c
i.e. c = 7a

The solutions x1 & x2 of the quadratic equation ax^2 + bx + c = 0 are standard :
x1 = [-b-sqroot(b^2-4ac)] / 2a
x2 = [-b+sqroot(b^2-4ac)] / 2a

and replacing b & c with their expressions (in terms of a) above gives us:
x1 = 1
x2 = 7



Coming to Q2, I dont understand the question, could you please elaborate

Another way of solving this :-
As per John, the 2 solutions of the quadratic equations are 6 and 2. the quadratic equation will be (x-6)(x-2)
i.e. x^2-8x+12=0 >>>we will take this as John's equation
As per Katy the solution is -7 and -1 and the quadratic equation takes the form (x+7)(x+1)=0
i.e.
x^2+8x+7=0>>>we will take this as Katy's equation.
As per the question"While John was solving the problem he took incorrect value for C, while katy took incorrect value of B"
Ignore the value of C John calculated i.e. 12 and ignore the value of B calcualted by Katy.
Combine the equations, the answer is x^2-8x+7=0 , i.e (1,7)

Regarding the 2nd equation,
"2.
x= a^r
y=b^s
z=c^t
a=2
b=3
c=4
r=3^4
s=4^3
c=2^4
then the relationship among x,y,z (eg:x>y>z) "
I think there should be a correction-Instead of c=2^4 it should be t=2^4