find the value of k so that the sum of the sum of the roots of the quadtric equation 3x^2+(2k+1)x-(k+4)=0 is equal to the products of the roots
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f(x)=3x^2+(2k+1)x-(k+4)=0
Here, a=3
b=(2k+1)
c= -(k+4)
ATQ
Alpha +Beta=Alpha×Beta
-b/a=c/a
-b=c
-(2k+1)=-(k+4)
2k+1=k+4
3k=3
k=1
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