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
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