Answer:
The given equation can be written as \[(k+1){{x}^{2}}-2(k-1)x+1=0\] Since the equation has equal roots \[4{{(k-1)}^{2}}-4(k+1)=0\] \[4({{k}^{2}}+1-2k)-4(k+1)=0\] \[4{{k}^{2}}+4-8k-4k-4=0\] \[4{{k}^{2}}-12k=0\] \[4k(k-3)=0\] \[k=0,3\] \[\therefore \] Non zero value of k is 3. And the equation becomes, \[4{{x}^{2}}-4x+1=0\] \[{{(2x-1)}^{2}}=0\] \[x=\frac{1}{2},\frac{1}{2}\] which are the required roots of the given equation.
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