URGENT ANS REQD PLZ

1- SACHIN ND RAHUL ATTEMPT TO SOLVE A QUAD EQN SACHIN MADE MISTAKE IN WRTING CNST TERM ENDED UP IN ROOT 4,3 RAHUL DONE MISTAKE IN WRITING DOWN COEFF OF X TO GET ROOT 3,2 WAT IS CORRECT ROOT FOR EQN

2- IF THE ROOTS OF EQN X2-BX+C =0 BE 2 CONSECUTIVE INTEGERS FIND B2- 4C

As sachin made a mistake in writing constant term , so he got roots as 4 ,3
So according to sachin the quadratic eqaution is  ( x-4) (x-3 ) = 0
Or x2 -7x +12 = 0
As in this equation only 12 is wrong , so middle term is -7

Rahul done a mistake in writing down the coeffiecient of x , got  3,2
Hence the equation acc to him is ( x-2)(x-3) = 0
x2 -5x +6 = 0
Here only -5 is wrong , so constant term is 6

Hence the correct equation is x2 -7x +6  = 0
Root of it are x = 6 , 1 


2) As the roots are consecutive ,so let n and n + 1 be the roots 
So (x -n ) ( x-n-1) = 0
x2 -nx -nx +n2 -x +n=0 

x2 -2nx -x +n2 +n = 0 

Comparing it with x 2 -bx +c =0 
we have b = -(2n+1)  and c = n2 +n 
So b2 -4ac =  4n2+1 + 4n -4n2-4n  = 1
 

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1) Let the eqn. be ax2 + bx + c.

Product of roots = c/a = 6 [from rahul].

Sum of roots = -b/a = 7 [from sachin].

So c=6a,b=-7a.

Hence eqn. will x2-7x+6=0.

So roots of this eqn is 1,6.

2) One (1).

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